Exercise
Q1.
Evaluate \(\lim_{x \rightarrow 3} x + 5\) limit:
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 3} x + 5 = 3 + 5 = 8\)
Q2.
Evaluate \(\lim_{x \rightarrow \pi}\left ( x – \frac{22}{7} \right )\)
Answers::
\(\lim_{x \rightarrow \pi} ( x – \frac{22}{7}) =( \pi – \frac{22}{7})\)
Q3.
Evaluate \(\lim_{r \rightarrow l} \pi r^{2}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{r \rightarrow l} \pi r^{2} = \pi (l)^{2} = \pi \)
Q4.
Evaluate \(\lim_{x + 4}\frac{4x + 6}{x – 3}\)
Answers::
\(\lim_{x + 4}\frac{4x + 6}{x – 3} = \frac{4 (4) + 6}{4 – 3} = \frac{16 + 6}{1} = \frac{22}{1} = 22\)
Q5.
Evaluate \(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1} = \frac{(-1)^{12} + (-1)^{7} + 1}{-1 – 1 } = \frac{1 – 1 + 1}{-2} = -\frac{1}{2}\)
Q6.
Evaluate \(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)Put x + 3 = y so that \(y \rightarrow 1 \; as \; x \rightarrow 0\)
Accordingly, \(\\ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = \lim_{y \rightarrow 1}\frac{y^{5} – 3}{y – 3} \\ = \lim_{y \rightarrow 1}\frac{y^{5} – 3^{5}}{y – 3} \\ = 5.3^{5 – 1} \\ = 405 \\ ∴ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = 405\)
Q7.
Evaluate \(\lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4}\)
Answers:: Based on the formulae given in Limits and Derivatives
At x = 2, the given rational function has the value of 0/ 0
\(\\ \lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4} = \lim_{x \rightarrow 2}\frac{ \left ( x – 2 \right ) \left ( 3x + 5 \right ) }{\left ( x – 2 \right ) \left ( x + 2 \right ) } \\ = \lim_{x \rightarrow 2} \frac{3x + 5}{x + 2} \\ = \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4}\)
Q8.
Evaluate \(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3}\)
Answers:: Based on the formulae given in Limits and Derivatives
At x = 2, the given rational function has the value of 0/ 0
\(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3} = \lim_{x \rightarrow 3}\frac{(x- 3)(x + 3)(x^{2} + 9)}{(x – 3)(2x + 1)} \\ = \lim_{x \rightarrow 3}\frac{(x + 3)(x^{2} + 9)}{2x + 1} \\ = \frac{(3 + 3)(3^{2} + 9)}{2(3) + 1} \\ = \frac{6 \times 18}{7} \\ = \frac{108}{7}\)
Q9.
Evaluate \(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0)+ 1} = b\)
Q10.
Evaluate \(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)At z = 1, the given function has the value of 0/ 0
Put \(z^{\frac{1}{6}} = x \; so \; that \; z\rightarrow 1 \; as \; x \rightarrow 1 \\ According to this, \; \lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}} – 1}{z^{^{\frac{1}{6}}} – 1} = \lim_{x \rightarrow 1} \frac{x^{2} – 1}{x – 1} \\ = \lim_{x \rightarrow 1}\frac{x^{2} – 1^{2}}{x – 1} \\ = 2.1^{2 – 1} \\ = 2 \\ ∴ \lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1} = 2\)
Q11.
Evaluate \(\lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a}, \; a + b + c \neq 0\)
Answers::
\(\\ \lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a} = \frac{a(1)^{2} + b(1) + c}{c(1)^{2} + b(1) + a}\\ = \frac{a + b + c}{a + b + c} \\ = 1\)
Q12.
\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)At x = -2, the given function has the value of 0/ 0
\(Now, \; \lim_{x \rightarrow 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \rightarrow -2} \frac{\left ( \frac{2 + x}{2x} \right )}{x + 2} \\ = \lim_{x \rightarrow -2}\frac{1}{2x} \\ = \frac{1}{2 (-2) } = \frac{-1}{4}\)
Q13.
Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)At x = 0, the given function has the value of 0/ 0
\(\\ Now, \lim_{x \rightarrow 0}\frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \frac{sin ax}{ax} \times \frac{ax}{bx} \\ = \lim_{x \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \times \left ( \frac{a}{b} \right ) \\ = \frac{a}{b} \lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \\ = \frac{a}{b} \times 1 \\ = \frac{a}{b}\)
Q14.
Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)
Answers::
\(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)At x = 0, the given function has the value of 0/ 0
\(\\ Now, \lim_{x \rightarrow 0}\frac{sin ax}{sin bx} = \lim_{x \rightarrow 0} \frac { \left (\frac{ \sin ax}{ax} \right ) \times ax}{\left ( \frac{\sin bx}{bx} \right ) \times bx}\\ = \left ( \frac{a}{b} \right ) \times \frac{\lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right )}{\lim_{bx \rightarrow 0} \left ( \frac{\sin bx}{bx} \right )} \\ = \left ( \frac{a}{b} \right ) \times \frac{1}{1} \\ = \frac{a}{b}\)
Q15.
Evaluate \(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)It seems \(x \rightarrow \pi \Rightarrow \left ( \pi – x \right ) \rightarrow 0\)
\(\\ ∴ \lim_{x \rightarrow \pi} \frac{\sin (\pi – x)}{\pi (\pi – x)} = \frac{1}{\pi}\lim_{\left ( z – x \right ) – 1}\frac{\sin(\pi – x)}{(\pi – x)} \\ = \frac{1}{\pi} \times 1 \\ = \frac{1}{\pi}\)
Q16.
Evaluate \(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x}\)
Answers:; Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x} = \frac{\cos 0}{\pi – 0} = \frac{1}{\pi}\)
Q17.
Evaluate \(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)At x = 0, the given function has the value of 0/ 0
\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\cos 2x – 1}{\cos x – 1} = \lim_{x \rightarrow 0}\frac{1 – 2 \sin^{2} x – 1}{1 – 2 \sin^{2} \frac{x}{2} – 1} \\ = \lim_{x \rightarrow 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}} = \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin^{2}x}{x^{2}} \right ) \times x^{2}}{\left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )\times \frac{x^{2}}{4}} \\ = 4 \frac{\lim_{x\rightarrow 0}\left ( \frac{\sin^{2}x}{x^{2}} \right )}{\lim_{x \rightarrow 0} \left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )}\) \(\\ = 4 \frac{\left (\lim_{x \rightarrow 0}\frac{\sin x}{x} \right )^{2}}{\left ( \lim_{\frac{x}{2} \rightarrow 0} \frac{sin \frac{x}{2}}{\frac{x}{2}} \right )^{2}} \\ = 4\frac{1^{2}}{1^{2}} \\ = 4\)
Q18.
Evaluate \(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)At x = 0, the given function has the value of 0/ 0
\(\\ Now \\ \lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x} = \frac{1}{b} \lim_{x \rightarrow 0}\frac{x(a + \cos x)}{\sin x}\\ = \frac{1}{b} \lim_{x \rightarrow 0} \left ( \frac{x}{\sin x} \right ) \times \lim_{x \rightarrow 0}(a + \cos x) \\ = \frac{1}{b} \times \frac{1}{\left ( \lim_{x \rightarrow 0} \frac{\sin x}{x} \right )} \times \lim_{x \rightarrow 0}\left ( a + \cos x \right ) \\ = \frac{1}{b} \times \left ( a + \cos 0 \right ) \\ = \frac{a + 1}{b}\)
Q19.
Evaluate \(\lim_{x \rightarrow 0} x \sec x\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 0} x \sec x = \lim_{x \rightarrow 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0\)
Q20.
Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} a, \;b \; \; a + b \neq 0\)
Answers:: Based on the formulae given in Limits and Derivatives
At x =0, the given function has the value of 0/ 0
\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} \\ \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin ax}{ax} \right )ax + bx}{ax + bx\left ( \frac{\sin bx}{bx} \right )} \\ =\) \(\\ = \frac{\left ( \lim_{ax \rightarrow 0} \frac{\sin ax}{ax} \right ) \times \lim_{x \rightarrow 0}\left ( ax \right ) + \lim_{x \rightarrow 0} bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0} bx\left ( \lim_{bx \rightarrow 0} \frac{\sin bx}{bx} \right )} \\ = \frac{\lim_{x \rightarrow 0}(ax) + \lim_{x \rightarrow 0}bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0}bx} \\ = \frac{\lim_{x \rightarrow 0}(ax + bx)}{\lim_{x \rightarrow 0}(ax + bx)} \\ = \lim_{x \rightarrow 0}(1) \\ = 1\)
Q21.
Evaluate \(\lim_{x \rightarrow 0}(\csc x – \cot x)\)
Answers:: Based on the formulae given in Limits and Derivatives
At x =0, the given function has the value of ∞ – ∞
\(\\ Now, \\ \lim_{x \rightarrow 0}(\csc x – \cot x) \\ = \lim_{x \rightarrow 0}\left ( \frac{1}{\sin x} – \frac{\cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\left ( \frac{1 – \cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\frac{\left (\frac{1 – \cos x}{\sin x} \right )}{\left (\frac{\sin x}{x} \right )} \\ = \frac{\lim_{x \rightarrow 0}\frac{1 – \cos x}{x}}{\lim_{x \rightarrow 0}\frac{\sin x}{x}} \\ = \frac{0}{1} \\ = 0\)
Q22.
\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)At \( x = \frac{\pi}{2}\), the given function has the value of 0/ 0
\(\\ Now, \; put \; x – \frac{\pi}{2} = y \; so \; that \; x \rightarrow \frac{\pi}{2}, \; y \rightarrow 0 \\ ∴ \lim_{x \rightarrow \frac{\pi}{2}}\frac{\tan2x}{x – \frac{\pi}{2}} = \lim_{y \rightarrow 0}\frac{tan 2\left ( y + \frac{\pi}{2} \right ) }{y} \\ = \lim_{y \rightarrow 0} \frac{tan \left ( \pi + 2y \right ) }{y} \\ =\lim_{y \rightarrow 0} \frac{\tan 2y}{y} \\ = \lim_{y \rightarrow 0}\frac{\sin 2y}{y \cos 2y} \\ = \lim_{y \rightarrow 0}\left ( \frac{\sin 2y}{2y} \times \frac{2}{\cos 2y} \right ) \\ = \left ( \lim_{2y \rightarrow 0} \frac{\sin 2y}{2y} \right ) \times \lim_{y \rightarrow 0}\left ( \frac{2}{\cos 2y} \right ) \\ =1 \times \frac{2}{\cos 0} \\ = 1 \times \frac{2}{1} \\ = 2\)
Q23.
Find \(\\ \lim_{x\rightarrow 0} f(x) \; and \; \lim_{x\rightarrow 0} f(x), \\ where f(x) = \left\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\right.\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function \(f(x) =\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\)
\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0}\left [ 2x + 3 \right ] = 2(0) + 3 = 3\)
\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} 3(x + 1) = 3(0 + 1) = 3\)
\(∴ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = 3\)
\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)
\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)
\(∴ \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x)\) = 6
Q24.
Find \(\lim_{x \rightarrow 1} f(x), \; where \; f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function is as follows:
\(f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 1^{-} f(x)} = \lim_{x \rightarrow 1}[x^{2} – 1] = 1^{2} – 1 = 1 – 1 = 0\) \(\lim_{x \rightarrow 1^{+} f(x)} = \lim_{x \rightarrow 1}[-x^{2} – 1] = -1^{2} – 1 = 1 – 1 = 0\)We know that \(\lim_{x \rightarrow 1^{-}} f(x) \neq \lim_{x \rightarrow 1^{+}} f(x)\)
Hence we have, \(\lim_{x \rightarrow 1} f(x)\) doesn’t exist.
25.
Evaluate \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function \(f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)
When \( \left | x \right | = -x \)
\(\lim_{x \rightarrow 0 ^{-}} f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{\left | x \right |}{x} \right ]\)= \(\lim_{x \rightarrow 0 \left ( \frac{ -x }{x} \right )}\)
= \(\lim_{x \rightarrow 0 \left ( -1 \right )}\)
= -1
When \( \left | x \right | = x \)
\(\lim_{x \rightarrow 0 ^{+}} f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{\left | x \right |}{x} \right ]\)= \(\lim_{x \rightarrow 0 \left [ \frac{ -x }{x} \right ]}\)
= \(\lim_{x \rightarrow 0 \left ( 1 \right )}\)
= 1
We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)
Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.
Q26.
Find \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{x}{\left | x \right |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)
Answers:: Based on the formulae given in Limits and Derivatives
f(x) = \(\{\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\)
When \(\left | x \right | = -x\)
\(\lim_{x \rightarrow 0^{-}}f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{x}{\left | x \right |} \right ]\) \( = \lim_{x \rightarrow 0}\left [ \frac{x}{ – x } \right ]\) \( = \lim_{x \rightarrow 0}\left ( – 1 \right )\)= -1
When \(\left | x \right | = x\)
\(\lim_{x \rightarrow 0^{+}}f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{x}{\left | x \right |} \right ]\) \( = \lim_{x \rightarrow 0}\left [ \frac{x}{ x } \right ]\) \( = \lim_{x \rightarrow 0}\left ( 1 \right )\)= 1
We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)
Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.
Q27.
Find \(\lim_{x \rightarrow 5} f(x), \; where \; f(x) = \left | x \right | – 5\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function is as follows: \( f(x) = \left | x \right | – 5\)
When \( x > 0, \left | x \right | = x \)
\(\lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}}[\left | x \right | – 5]\)= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)
= 5 – 5
= 0
When \( x > 0, \left | x \right | = x \)
\(\lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}}(\left | x \right | – 5)\)= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)
= 5 – 5
= 0
\(∴ \lim_{x \rightarrow 5^{-}}f(x) = \lim_{x \rightarrow 5^{+}}f(x) = 0\)Hence, \(\lim_{x \rightarrow 5 } f(x) = 0 \)
Q28.
Suppose \(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\)
And \(\lim_{x \rightarrow 1} f(x) = f(1)\) what can be the values of b and a?
Answers:: Based on the formulae given in Limits and Derivatives
Given function
\(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(a + bx) = a + b\) \(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(b – ax) = b – a \)f(1) = 4
Given that \(\lim_{x \rightarrow 1} f(x) = f(1)\).
\(∴ \lim_{x \rightarrow 1^{-}}f(x) = \lim_{x \rightarrow 1^{+}}f(x) = \lim_{x \rightarrow 1} f(x) = f(1)\)=> a+ b = 4 and b – a = 4
Solving both , we get a = 0 and b = 4.
Hence proved that, b and a are 4 and 0
Q29.
A function \(f(x) = (x – a_{1})(x – a_{2}) . . . . (x – a_{n})\) is define by fixed real numbers a1, a2, . . . . an.
Find \(\lim_{x \rightarrow a_{1}}f(x)\). For some \(a \neq a_{1}, a_{2}, . . . . , a_{n} \; compute \; \lim_{x \rightarrow a_{n}} f(x)\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function = \(f(x) = (x – a_{1})(x – a_{2}) . . . (x – a_{n})\)
\(\lim_{x \rightarrow a_{1}}f(x) = \lim_{x \rightarrow a_{1}}[(x – a_{1})(x – a_{2}) . . . (x – a_{n})]\)\(= [\lim_{x \rightarrow a_{1}}(x – a_{1})][\lim_{x \rightarrow a_{1}}(x – a_{2})] . . . [\lim_{x \rightarrow a_{1}}(x – a_{n})]\)
\(= \left ( a_{1} – a_{1} \right ) \left ( a_{1} – a_{2} \right ) . . . . (a_{1} – a_{n}) = 0\)
\(∴ \lim_{x \rightarrow a_{1}]}f(x) = 0\)
Now, \(\lim_{x \rightarrow a_{1}]}f(x) = \lim_{x \rightarrow a}[\left (x – a_{1} \right )\left ( x – a_{2} \right ) . . . \left ( x – a_{n} \right )]\)
= \(= \lim_{x \rightarrow a}[ x – a_{1} ] [ x – a_{2}] . . . [x – a_{n}]\)
\(( a_{1} – a_{1}) ( a_{1} – a_{2} ) . . . . (a_{1} – a_{n})\)
\( ∴ \lim_{x \rightarrow a} f(x) =( a_{1} – a_{1})( a_{1} – a_{2}) . . . . (a_{1} – a_{n}) \)
Q30.
If \(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)
\(\lim_{x \rightarrow a} f(x)\) will exist for what values?
Answers:: Based on the formulae given in Limits and Derivatives
Given function:
\(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)When a = 0
\(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0^{-}}\left ( \left | x \right | + 1 \right )\)\(= \lim_{x \rightarrow 0}\left ( – x + 1 \right )\) [if x < 0, |x| = -x]
= -0 + 1
= 1
\(\lim_{x \rightarrow 0^{+} f(x)} = \lim_{x \rightarrow 0^{+}}\left ( \left | x \right | – 1 \right )\)\(= \lim_{x \rightarrow 0}\left ( x – 1 \right )\) [if x > 0, |x| = x]
= 0 – 1
= -1
We have, \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)
\(∴ \lim_{x \rightarrow 0} f(x)\) doesn’t exist.
When a < 0
\(\lim_{x \rightarrow a^{-}}f(x) = \lim_{x \rightarrow a^{-}}(\left | x \right | + 1)\) \(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)= -a + 1
\(\lim_{x \rightarrow a^{+}}f(x) = \lim_{x \rightarrow a^{+}}(\left | x \right | + 1)\) \(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)= – a + 1
\(∴ \lim_{x \rightarrow a^{-} } = \lim_{x \rightarrow a^{+} } = – a + 1\)Hence, at x = a limit of f(x) exist, where a > 0
Thus, \(\lim_{x \rightarrow a} f(x) \; exist \; for \; all \; a \neq 0\)
Q31.
Evaluate \(\lim_{x \rightarrow 1} f(x)\)
If f(x) satisfies, \(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\)
Answers:: Based on the formulae given in Limits and Derivatives
\(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\) \( \Rightarrow \frac{\lim_{x \rightarrow 1} \left ( f(x) – 2 \right )}{ \lim_{x \rightarrow 1} \left ( x^{2} – 1 \right )} = \pi\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi \lim_{x \rightarrow 1} (x^{2} – 1)\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi (1^{2} – 1)\) \(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = 0 \) \(\Rightarrow \lim_{x \rightarrow 1} f(x) – \lim_{x \rightarrow 1} 2 = 0 \) \(\Rightarrow \lim_{x \rightarrow 1} f(x) – 2 = 0 \) \(∴ \lim_{x \rightarrow 1} f(x) = 2\)
Q32.
If \(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\)
For what values of a and b does \(\lim_{x \rightarrow 0} f(x) \; and \; \lim_{x \rightarrow 1} f(x)\) exist?
Answers:: Based on the formulae given in Limits and Derivatives
Given function
\(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\) \(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0}(ax^{2} + b)\)= a(0)2 + b
= b
\(\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0}(bx + a)\)= b(1) + a
= a + b
\(∴ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} f(x)\)Thus we get, for integral values of n and m \(\lim_{x \rightarrow 1}f(x)\) exist.
Exercise-13.2
Q1. Find derivative for x2 – 2 at x = 10
Answers:: Based on the formulae given in Limits and Derivatives
Let f(x) = x2 – 2
Accordingly,
\(f'(10) = \lim_{h \rightarrow 0} \frac {f(10 + h) – f(10)}{h}\)\(= \lim_{h \rightarrow 0} \frac {[(10 + h)^{2} – 2] – (10^{2} – 2)}{h}\)
\(= \lim_{h \rightarrow 0} \frac {10^{2} + 2. 10. h + h^{2} – 2 – 10^{2} + 2}{h}\)
\(= \lim_{h \rightarrow 0} \frac {20h + h^{2}}{h} \)
\(= \lim_{h \rightarrow 0} (20 + h) = (20 + 0) = 0 \)
Hence we have the, derivative for x2 – 2 at x = 10 is 20
Q2. Find derivative
99x at x = 100
Answers:: Based on the formulae given in Limits and Derivatives
Let f(x) = 99x,
Accordingly,
\(f'(100) = \lim_{h \rightarrow 0} \frac{f(100 + h) – f(100)}{h}\)\( = \lim_{h \rightarrow 0} \frac{99(100 + h) – 99(100)}{h}\)
\( = \lim_{h \rightarrow 0} \frac{99 \times 100 + 99h – 99 \times 100}{h}\)
\( = \lim_{h \rightarrow 0} \frac{ 99h }{h}\)
\( = \lim_{h \rightarrow 0} (99h) = 99 \)
Hence, derivative for 99x at x = 100 is 99
Q3. Find derivative
x at x = 1
Answers:: Based on the formulae given in Limits and Derivatives
Let f(x) = x
Accordingly,
\(f'(1) = \lim_{h \rightarrow 0} \frac{f(1 + h) – f(1)}{h}\)\(= \lim_{h \rightarrow 0} \frac{(1 + h) – 1}{h}\)
\(= \lim_{h \rightarrow 0} \frac{h}{h}\)
\(= \lim_{h \rightarrow 0} (1)\)
= 1
Hence, derivative for x at x = 1 is 1
Q4. Using first principle find derivative
(i) x3 – 27
(ii) (x – 1) (x – 2)
(iii) \(\frac{1}{x^{2}}\)
(iv) \(\frac{x + 1}{x – 1}\)
Answers:: Based on the formulae given in Limits and Derivatives
(i) Let f(x) = x3 – 27
From first principle,
\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)\( = \lim_{h \rightarrow 0}\frac{ [(x + h)^{3} – 27] – (x^{3} – 27)} {h} \)
\( = \lim_{h \rightarrow 0}\frac{ x^{3} + h^{3} + 3x^{2}h + 3xh^{2} – x^{3}}{h} \)
\( = \lim_{h \rightarrow 0}\frac{ h^{3} + 3x^{2}h + 3xh^{2}} {h} \)
\( = \lim_{h \rightarrow 0} \left ( h^{2} + 3x^{2} + 3xh \right ) \)
= 0 + 3x2 + 0 = 3x2
(ii) Let f(x) = (x – 1)(x – 2)
From first principle,
\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0}\frac{(x + h – 1) (x + h – 2) – (x – 1) (x – 2)} {h}\)
\(= \lim_{h \rightarrow 0}\frac{(x^{2} + hx – 2x + hx + h^{2} – 2h – x – h + 2 ) – (x^{2} – 2x – x + 2)} {h}\)
\(= \lim_{h \rightarrow 0}\frac{(hx + hx + h^{2} – 2h -h )} {h}\)
\(= \lim_{h \rightarrow 0}\frac{( 2hx + h^{2} – 3h )} {h}\)
\(= \lim_{h \rightarrow 0} ( 2x + h – 3 )\)
\( ( 2x + h – 3 )\)
= 2x – 3
(iii) Let \(f(x) = \frac{1}{x^{2}}\)
From first principle
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\( = \lim_{h \rightarrow 0} \frac{\frac{1}{(x + h)^{2}} – \frac{1}{x^{2}}}{h}\)
\( = \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{ x^{2} – (x + h )^{2}}{x^{2} (x + h )^{2}}\right ]\)
\( = \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{ x^{2} – x^{2} – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)
\( = \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{ – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)
\( = \lim_{h \rightarrow 0} \left [\frac{ – h – 2hx}{x^{2} (x + h )^{2}}\right ]\)
\(\frac{0 – 2x}{x^{2}(x + 0)^{2}} = \frac{-2}{x^{3}}\)
(iv) Let \(f(x) = \frac{x +1}{x – 1}\)
From first principle,
\( = \lim_{h \rightarrow 0} \frac{\left ( \frac{x + h + 1}{x + h – 1} – \frac{x + 1}{x – 1} \right )}{h}\)\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x – 1 \right ) \left ( x+ h + 1 \right ) – \left ( x + 1 \right ) \left ( x + h – 1 \right )}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)
\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x^{2} + hx + x – x – h – 1 \right ) – \left ( x^{2} + hx – x + x + h – 1 \right )}{\left ( x- 1 \right )\left ( x + h + 1 \right )} \right ]\)
\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{-2h}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)
\( = \lim_{h \rightarrow 0} \left [ \frac{-2}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)
\(= \frac{-2}{\left ( x – 1 \right )\left ( x – 1 \right )} = \frac{-2}{\left ( x – 1 \right )^{2}}\)
Q5. Prove \(f'(1) = 100 f'(0)\)
For the function \(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)
Answers:: Based on the formulae given in Limits and Derivatives
Given function
\(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)\(\frac{d}{dx} f(x) = \frac{d}{dx}\left [ \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1 \right ]\)
\(\frac{d}{dx} f(x) = \frac{d}{dx} \left ( \frac{x^{100}}{100} \right ) + \frac{d}{dx}\left ( \frac{x^{99}}{99} \right ) + … + \frac{d}{dx}\left ( \frac{x^{2}}{2} \right ) + \frac{d}{dx}\left ( x \right )+ \frac{d}{dx}\left ( 1 \right )\)
Using theorem \(\frac{d}{dx} \left ( x^{n} \right ) = n x^{n – 1}, \) we get
\(\frac{d}{dx} f(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + . . . . + \frac{2x}{2} + 1 + 0\)
\(= x^{99} + x^{98} + …. + x + 1\)
∴ \(\frac{d}{dx} f(x) = x^{99} + x^{98} + …. + x + 1\)
At x = 0
f'(0) = 1
At x = 1,
f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100
Hence, f’(1) = 100 x f1 (0)
Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an
Answers:: Based on the formulae given in Limits and Derivatives
Let \(f(x) = x ^{n} + ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n}\)
\(f’(x) = \frac{d}{dx} \left ( x ^{n} + ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n} \right ) \)
\(= \frac{d}{dx} \left ( x ^{n} \right ) + a\frac{d}{dx} \left ( x ^{n – 1} \right ) + a^{2} \frac{d}{dx} \left ( x ^{n – 2} \right ) + . . . + a^{n – 1}\frac{d}{dx} \left ( x \right ) + a^{n} \frac{d}{dx} \left ( 1 \right )\)
Using \(\frac{d}{dx}x^{n} = nx^{n -1}\), we have
\(f'(x) = nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1} + a^{n}(0)\)
\(= nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1}\)
Q7. Find the derivative for
(i) (x – m)(x – n)
(ii) (ax2 + b)2
(iii) \(\frac{x – a}{x – b}\)
Answers::
(i) Let f(x) = (x – m)(x – n)
\(\Rightarrow f(x) = x^{2} – (m + n) x + mn\)
\(f'(x) = \frac{d}{dx}(x^{2} – (m + n)x + mn)\)
\(= \frac{d}{dx}(x^{2}) – (m + n) \frac{d}{dx}(x) + \frac{d}{dx}(mn)\)
Using \(= \frac{d}{dx}(x^{n}) = nx^{n – 1}\), we get
\(f'(x) = 2x -(m + n) + 0 = 2x – m – n\)
(ii) Let f(x) = (ax2 + b)2
\(\Rightarrow f(x) = a^{2}x^{2} + 2abx + b^{2}\)
\(f'(x) = \frac{d}{dx} (a^{2}x^{4} + 2abx^{2} + b^{2}) = a^{2}\frac{d}{dx}(x^{4}) + 2ab\frac{d}{dx}(x^{2}) + \frac{d}{dx}(b^{2})\)
Using \(\frac{d}{dx}x^{n} = nx^{n – 1}\), we have
\(f’\left ( x \right ) = a ^{2}(4x^{3}) + 2ab (2x) + b^{2}(0)\)
= 4a2x3 + 4abx
= 4ax(ax2 + b)
(iii) Let \(f(x) = \frac{(x – a)}{(x – b)}\)
\(f'(x) = \frac{d}{dx} \left (\frac{x – a}{x – b} \right )\)
Quotient rule,
\(f'(x) = \frac {(x – b)\frac{d}{dx}(x – a) – (x – a)\frac{d}{dx}(x- b)}{(x – b)^{2}}\)\(= \frac {(x – b)(1) – (x – a)(1)}{(x – b)^{2}}\)
\(= \frac {x – b – x + a}{(x – b)^{2}}\)
\(= \frac { a – b }{(x – b)^{2}}\)
Q8. If a is constant find derivative of \(\frac {x^{n} – a^{n}}{x – a}\)
Answers::
Let \(\frac {x^{n} – a^{n}}{x – a}\)
\(\Rightarrow f'(x) = \frac{d}{dx}\left ( \frac {x^{n} – a^{n}}{x – a} \right )\)
By Question rule
\(\Rightarrow f'(x) = \frac{(x – a) \frac{d}{dx}(x^{n} – a^{n}) – (x^{n} – a^{n})\frac{d}{dx}(x – a)}{(x – a)^{2}}\)\(= \frac{(x -a)(nx^{n – 1} – 0) – (x^{n} – a^{n})}{(x – a)^{2}}\)
\(= \frac{nx^{n} – anx^{n – 1} – x^{n} + a^{n}}{(x – a)^{2}}\)
Q9.
(i) \( 2x – \frac {3}{4} \)
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x-3 (5 + 3x)
(iv) x5 (3 – 6x-9)
(v) x-4 (3 – 4x-5)
(vi) \(= \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)
Answers:. Based on the formulae given in Limits and Derivatives
(i) Let \( f(x) = 2x – \frac {3}{4} \)
\(f'(x) = \frac{d}{dx}\left ( 2x – \frac{3}{4} \right )\)
\(= 2\frac{d}{dx}\left ( x \right ) – \frac {d}{dx} \left ( \frac{3}{4} \right )\)
= 2 – 0
= 2
(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)
By Leibnitz product rule, Based on the formulae given in Limits and Derivatives
\(f'(x) = (5x^{3} + 3x – 1) \frac{d}{dx}(x – 1) + (x – 1) \frac{d}{dx}(5x^{3} + 3x – 1 )\)
= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)
= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)
= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3
= 20x3 – 15x2 + 6x – 4
(iii) Let f(x) = x-3 (5 + 3x)
\(f'(x) = x^{-3}\frac{d}{dx}(5 + 3x) + (5 + 3x)\frac{d}{dx}(x^{-3})\)
\( = x^{-3}(0 + 3) + (5 + 3x) (-3x^{-3 – 1} )\)
\( = x^{-3}( 3) + (5 + 3x) (-3x^{-4} )\)
= 3x-3 – 15x-4 – 9x-3
= -6x-3 – 15x-4
\(= -3 x ^{-3} \left ( 2 + \frac{5}{x} \right )\)
\(= \frac{-3 x ^{-3} }{x}\left ( 2x + 5 \right )\)
\(= \frac{-3 }{ x ^{4} }\left ( 2x + 5 \right )\)
(iv) let f(x) = x5 (3 – 6x-9)
From Leibnitz product rule, Based on the formulae given in Limits and Derivatives
\(f'(x) = x^{5} \frac{d}{dx}(3 – 6x^{-9}) + (3 – 6x^{-9}) \frac{d}{dx}(x^{5})\)\(= x^{5} \left \{ 0 – 6 (-9)x^{-9 – 1} \right \} + (3 – 6x^{-9})(5x^{4})\)
\(= x^{5} (54x ^{-10}) + 15x^{4} – 30x^{-5}\)
= 54x-5 + 15x4 – 30x-5
= 24x-5 + 15x4
= \(= 15x^{4} + \frac{24}{x^{5}}\)
(v) Let f(x) = x-4 (3 – 4x-5)
From Leibnitz product rule, Based on the formulae given in Limits and Derivatives
\(f'(x) = x^{-4}\frac{d}{dx}\left ( 3 – 4x^{-5} \right ) + (3 – 4 x^{-5})\frac{d}{dx}(x^{-4})\)\(= x^{-4}\left \{ 0 – 4 (-5)x^{-5 – 1} + (3 – 4x^{-5})(-4)x^{-4 -1} \right \}\)
= x-4 (20x-6) + (3 – 4x-5)(-4x-5)
= 36x-10 – 12x-5
\(= -\frac{12}{x^{5}} + \frac{36}{x^{10}}\)
(iv) Let \( f(x) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)
\(f'(x) = \frac{d}{dx}\left ( \frac{2}{x + 1} \right ) – \frac{d}{dx} \left ( \frac{x^{2}}{3x – 1} \right )\)
From quotient rule Based on the formulae given in Limits and Derivatives,
\(f'(x) = \left [ \frac{\left ( x + 1 \right ) \frac{d}{dx}(2) – 2 \frac{d}{dx} (x + 1) }{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1) \frac{d}{dx} (x^{2}) – x^{2} \frac{d}{dx}\left ( 3x – 1 \right )}{(3x – 1)^{2}} \right ]\)\(= \left [ \frac{(x + 1)(0) – 2(1)}{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1)(2x) – (x^{2})(3)}{(3x – 1)^{2}} \right ]\)
\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{6x^{2} – 2x – 3x^{2}}{(3x – 1)^{2}} \right ]\)
\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{3x^{2} – 2x} {(3x – 1)^{2}} \right ]\)
\(= \frac{-2}{(x + 1)^{2}} – \frac{x(3x – 2)} {(3x – 1)^{2}}\)
Q10. Using first principle find derivative of \(\cos x\)
Answers:: Based on the formulae given in Limits and Derivatives
Let \(f(x) = \cos x\).
According to first principle we have
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0} \frac{\cos (x + h) – \cos x}{h}\)
\(= \lim_{h \rightarrow 0} \left [\frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \right ]\)
\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h) – \sin x \sin h }{h} \right ]\)
\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h)}{h} – \frac{\sin x \sin h }{h} \right ]\)
\(=- \cos x \left ( \lim_{h \rightarrow 0} \frac{1 – \cosh}{h} \right ) – \sin x\lim_{h \rightarrow 0}\left ( \frac{sin h}{h} \right )\)
= \(= – \cos x (0) – \sin x (1)\) \(\left [ \lim_{h \rightarrow 0} \frac{1 – cos h}{h} = 0 \; and \; \lim_{h \rightarrow 0}\frac{\sin h }{h} = 1 \right ]\)
= – sin x
f’(x) = – sin x
Q11.
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7
(vii) 2tan x – 7sec x
Answers::
(i) Let f’(x) = sin x cos x
According to first principle, Based on the formulae given in Limits and Derivatives
\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) -f(x)}{h}\)\(= \lim_{h \rightarrow 0}\frac{\sin (x + h) \cos(x + h) – \sin x \cos x}{h}\)
\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2\sin (x + h) \cos (x + h) – 2 \sin x cos x \right ]\)
\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ \sin 2(x + h) – \sin 2x \right ]\)
\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2 \cos \frac{2x + 2h + 2x}{2} . \sin \frac{2x + 2h – 2x}{2} \right ]\)
\(= \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos \frac{ 4x + 2h }{2} \sin \frac{ 2h }{2} \right ]\)
\(\lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (2x + h ) \sin h \right ]\)
\(\lim_{h \rightarrow 0} \cos (2x + h ) \lim_{h \rightarrow 0} \frac{\sin h}{h}\)
= cos(2x + 0). 1
= cos 2x
(ii) Let f(x) = sec x
According to first principle,
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0} \frac{\sec (x + h) – \sec x}{h}\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos(x + h)}{\cos x \cos(x + h)} \right ]\)
\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \right ) \sin \left ( \frac{x – x – h}{2} \right )}{\cos (x + h)} \right ]\)
\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{2x + h}{2} \right ) \sin \left ( \frac{- h}{2} \right )}{\cos (x + h)} \right ]\)
\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\left [ \sin \left ( \frac{2x + h}{2} \right ) \frac{\sin \left ( \frac{h}{2}\right )}{\left (\frac{h}{2}\right )} \right ]}{\cos (x + h)}\)
\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}. \lim_{h \rightarrow 0}\frac{\sin\left ( \frac{2x + h}{2} \right ) }{\cos (x + h)}\)
\(= \frac{1}{\cos x} . 1 . \frac{\sin x}{ \cos x}\)
\(= \sec x \tan x\)
(iii)
Let f(x) = 5 sec x + 4 cos x
According to first principle,
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(\lim_{h \rightarrow 0}\frac{5\sec (x + h) + 4 \cos(x + h) – \left [ 5 \ sec x + 4 \ cos x \right ]}{h}\)
\(= 5\lim_{h \rightarrow 0}\frac{[\sec(x + h) – \sec x]}{h} + 4 \lim_{h \rightarrow 0}\frac{[\cos(x + h) – \cos x]}{h}\)
\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)
\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{\cos x – \cos (x + h)}{\cos x \cos (x + h)} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h}\left [ \cos x \cos h – \sin x \sin h – \cos x \right ]\)
\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{x + x + h}{2} \right )\sin \left ( \frac{x – x – h}{2} \right )}{cos(x + h)} \right ] + 4 \lim_{h\rightarrow 0}\frac{1}{h} [-\cos x (1 – \cos h) – \sin x sin h]\)
\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{ 2x + h}{2} \right )\sin \left ( \frac{ – h}{2} \right )}{cos(x + h)} \right ] + 4\left [ – \cos x \lim_{h \rightarrow 0} \frac{(1 – \cos h)}{h} – \sin x \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ] \)
\(= \frac{5}{\cos x}. \lim_{h \rightarrow 0} \left [ \frac{\sin\left ( \frac{2x + h}{2} \right ). \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}}{\cos (x + h)}\right ] + 4 [(-\cos x).(0) – (\sin x). 1]\)
\(= \frac{5}{\cos x}. \left [ \lim_{h \rightarrow 0} \frac{\sin\left ( \frac{2x + h}{2} \right )}{\cos (x + h)}. \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ] + 4 \sin x\)
\(= \frac{5}{\cos x}. \frac{\sin x}{\cos x}. 1 – 4 \sin x\)
\(= 5 \sec x \tan x – 4 \sin x\)
(iv)
Let f(x) = cosec x
According to first principle
\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)\(f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ] \)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin(x + h)}{ \sin (x + h) \sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ). \sin \left ( \frac{x – x – h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ). \sin \left ( \frac{- h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )} }{\sin (x + h) . \sin x}\)
\(= \lim_{h \rightarrow 0} \left ( \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). }{\sin (x + h) . \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)
\(= \left ( \frac{-\cos x}{\sin x \sin x} \right ).1\)
= – cosec x cot x
(v)
Let f(x) = 3cot x + 5cosec x
According to first principle,
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0} \frac{3 \cot (x + h) + 5 \csc (x + h) – 3 \cot x – 5 \csc x}{h}\)
\(= 3 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] + 5 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \csc x]\) ….. (1)
Now,
\( \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] \)
= \( \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos (x + h)}{\sin (x + h)} – \frac{\cos x}{\sin x} \right ] \)
= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos(x + h)\sin x – \cos x \sin (x + h)}{\sin x sin(x + h)} \right ]\)
= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x – x – h)}{\sin x sin(x + h)} \right ]\)
= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (- h)}{\sin x sin(x + h)} \right ]\)
= \(– \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ). \left ( \lim_{h \rightarrow 0} \frac{1}{\sin x. \sin (x + h)} \right )\)
= \(– 1 . \frac{1}{\sin x . \sin (x + 0)} = \frac{-1}{\sin^{2}x} = -\csc ^{2}x\) ……. (2)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin (x + h) }{\sin x \sin (x + h)} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ) . \sin \left ( \frac{x – x – h}{2} \right )}{\sin (x + h) \sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ) . \sin \left ( \frac{- h}{2} \right )}{\sin (x + h) \sin x} \right ]\)
\(= \lim_{h \rightarrow 0} \frac{ -\cos \left ( \frac{ 2x + h}{2} \right ) . \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}}{\sin (x + h) \sin x}\)
\(= \lim_{h \rightarrow 0} \left ( \frac{ -\cos \left ( \frac{ 2x + h}{2} \right )} {\sin (x + h) \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)
\(= \left ( \frac{- \cos x }{\sin x \sin x} \right ) . 1\)
\(= – \csc x \cot x\) ………. (3)
From eqn (1), (2), and (3), we obtain
\(f'(x) = – 3 \csc^{2} x – 5 \csc x \cot x\)
(vi)
Let f(x) = 5sin x – 6cos x + 7
According to first principle,
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \sin(x + h) – 6 \cos (x + h) + 7 – 5 \sin x + 6 \cos x – 7 \right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \left \{ \sin(x + h) – \sin x \right \} – 6 \left \{ \cos (x + h) – \cos x \right \} \right ]\)
\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sin(x + h) – \sin x \right ] – 6 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)
\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{x + x + h}{2} \right ) \sin \frac{x + h – x}{2} \right ] – 6 \lim_{h \rightarrow 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h}\)
\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{ 2x + h}{2} \right ) \sin \frac{ h }{2} \right ] – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h ) – \sin x \sin h }{h} \right ] \)
\(= 5 \lim_{h \rightarrow 0} \left ( \cos\left ( \frac{ 2x + h}{2} \right ) \frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ) – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h )}{h} – \frac {\sin x \sin h }{h} \right ]\)
\(= 5 \left [ \lim_{h \rightarrow 0} \cos\left ( \frac{ 2x + h}{2} \right ) \right ] \left [ \lim_{\frac{h}{2} \rightarrow 0}\frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ] – 6 \left [ (- \cos x ) \left ( \lim_{h \rightarrow 0} \frac{ (1 – \cos h )}{h} \right ) – \sin x \lim_{h \rightarrow 0} \left ( \frac {\sin h }{h} \right ) \right ]\)
\(= 5 \cos x. 1 – 6 [(-\cos x). (0) – \sin x.1]\)
= 5cos x + 6 sin x
(vii)
Let f (x) = 2 tan x – 7 sec x
Accordingly to first principle,
\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \tan (x + h) – 7 \sec (x + h) – 2 \tan x + 7 \sec x\right ]\)
\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \left \{ \tan (x + h) – \tan x \right \} – 7 \left \{ \sec (x + h) – \sec x \right \} \right ]\)
\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \tan (x + h) – \tan x \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sec (x + h) – \sec x \right ]\)
\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h)}{\cos (x + h)} – \frac{\sin x}{\cos x} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)
\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h) \cos x – \sin x \cos (x + h) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos (x + h) }{\cos x \cos (x + h)} \right ]\)
\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{ \sin (x + h – x) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \sin \left ( \frac{x – x – h}{2} \right ) \right )}{\cos x \cos (x + h)} \right ]\)
\(= 2 \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ) \left ( \lim_{h \rightarrow 0} \frac{ 1 }{ \cos x \cos (x + h)} \right ) – 7 \left( \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ) \left ( \lim_{h \rightarrow 0} \frac{ \sin \left ( \frac{ 2x + h}{2} \right ) }{\cos x \cos (x + h)} \right )\)
\(= 2 . 1 . \frac{1}{\cos x \cos x} – 7 . 1 \left ( \frac{\sin x}{\cos x \cos x} \right )\)
\(= 2 \sec^{2} x – 7 \sec x \tan x\)