Class 11: Limits – Exercise 29.8 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note:

$\displaystyle \lim \limits_{x \to a } \ f(x) = \lim \limits_{h \to 0 }\ f(a-h)$

Evaluate the following limits

$\displaystyle \text{ Question 1: } \lim \limits_{x \to \frac{\pi}{2} } \ \Big( \frac{\pi}{2} -x \Big) \tan x$

Answer:

$\displaystyle \text{Let } y = \frac{\pi}{2} -x$

$\displaystyle \text{As } x \rightarrow \frac{\pi}{2}, y \rightarrow 0$

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \Big( \frac{\pi}{2} -x \Big) \tan x$

$\displaystyle = \lim \limits_{y \to 0 } \ y \tan \Big( \frac{\pi}{2} -y \Big)$

$\displaystyle = \lim \limits_{y \to 0 } \ y \Big( \frac{\cos y}{\sin y} \Big)$

$\displaystyle = \lim \limits_{y \to 0 } \ \frac{\cos y}{\frac{\sin y}{y} }$

$\displaystyle = 1$

$\\$

$\displaystyle \text{ Question 2: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sin 2x}{\cos x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sin 2x}{\cos x} = \lim \limits_{x \to \frac{\pi}{2} } \ \frac{2 \sin x \cos x}{\cos x} = \lim \limits_{x \to \frac{\pi}{2} } \ 2 \sin x = 2$

$\\$

$\displaystyle \text{ Question 3: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\cos^2 x}{1- \sin x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\cos^2 x}{1- \sin x} = \lim \limits_{x \to \frac{\pi}{2} }\frac{1 - \sin^2 x}{1 - \sin x} = \lim \limits_{x \to \frac{\pi}{2} }\frac{(1-\sin x)(1+\sin x)}{(1-\sin x)} \\ \\ \\ { \hspace{6.0cm} = \lim \limits_{x \to \frac{\pi}{2} }1+\sin x = 1 + 1 = 2 }$

$\\$

$\displaystyle \text{ Question 4: } \lim \limits_{x \to \frac{\pi}{3} } \ \frac{\sqrt{1-\cos 6x}}{\sqrt{2}(\frac{\pi}{3}-x)}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{3} } \ \Bigg[ \frac{\sqrt{1-\cos 6x}}{\sqrt{2}(\frac{\pi}{3}-x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{3} } \ \Bigg[ \frac{\sqrt{2 \sin^2 3x}}{\sqrt{2}(\frac{\pi}{3}-x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{3} } \ \Bigg[ \frac{\sqrt{2} \sin 3x}{\sqrt{2}(\frac{\pi}{3}-x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{3} } \ \Bigg[ \frac{\sin 3x}{(\frac{\pi}{3}-x)} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sin 3(\frac{\pi}{3}+h)}{(\frac{\pi}{3}-(\frac{\pi}{3}+h))} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sin (\pi+3h)}{-h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{-\sin 3h}{-h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sin 3h}{3h} \times 3 \Bigg]$

$\displaystyle = 3$

$\\$

$\displaystyle \text{ Question 5: } \lim \limits_{x \to a } \ \frac{\cos x - \cos a}{x - a}$

Answer:

$\displaystyle \lim \limits_{x \to a } \ \frac{\cos x - \cos a}{x - a} = \lim \limits_{x \to a } \ \frac{-2 \sin (\frac{x+a}{2}) \sin (\frac{x-a}{2}) }{2 (\frac{x-a}{2} )} = \lim \limits_{x \to a } \ -\sin \Big( \frac{x+a}{2} \Big) \\ \\ \\ { \hspace{8.5cm} = -\sin \Big( \frac{2a}{2} \Big) = - \sin a }$

$\\$

$\displaystyle \text{ Question 6: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \tan x}{x - \frac{\pi}{4}}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \tan x}{x - \frac{\pi}{4}} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{1 - \tan ( \frac{\pi}{4} +h)}{(\frac{\pi}{4}+h- \frac{\pi}{4})} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{1 - (\frac{1+\tan h}{1 - \tan h} ) }{h} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{1 - \tan h - 1 - \tan h}{(1-\tan h)h} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{-2\tan h}{h(1-\tan h)} = \frac{-2}{1-0} = -2$

$\\$

$\displaystyle \text{ Question 7: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{1 - \sin x}{\Big( \frac{\pi}{2} - x \Big)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{1 - \sin x}{( \frac{\pi}{2} - x )^2} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{1 - \sin (\frac{\pi}{2} - h)}{\Big( \frac{\pi}{2} - (\frac{\pi}{2} -h ) \Big)^2} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{1- \cos h}{h^2} \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{h}{2}}{h^2 } \\ \\ \\ = \lim \limits_{h \to 0 } \ \frac{2}{4} \Bigg( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \Bigg)^2 \\ \\ \\ = \frac{1}{2}$

$\\$

$\displaystyle \text{ Question 8: } \lim \limits_{x \to \frac{\pi}{3} } \ \frac{\sqrt{3} - \tan x}{\pi - 3x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{3} } \ \Bigg[ \frac{\sqrt{3} - \tan x}{\pi - 3x} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sqrt{3} - \tan (\frac{\pi}{3} - h) }{\pi - 3(\frac{\pi}{3} - h)} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sqrt{3} - \Big( \frac{\tan \frac{\pi}{3} - \tan h}{1 + \tan \frac{\pi}{3} \tan h} \Big)}{\pi - 3( \frac{\pi}{3} - h)} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sqrt{3} - \Big( \frac{\sqrt{3} - \tan h}{1 + \sqrt{3} \tan h} \Big)}{3h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{\sqrt{3} + 3 \tan h - \sqrt{3} + \tan h }{( 1 + \sqrt{3} \tan h) 3 h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{4 \tan h }{3h ( 1 + \sqrt{3} \tan h) } \Bigg]$

$\displaystyle = \frac{4}{3}$

$\\$

$\displaystyle \text{ Question 9: } \lim \limits_{x \to a } \ \frac{a \sin x - x \sin a}{ax^2 - xa^2}$

Answer:

$\displaystyle \lim \limits_{x \to a } \ \frac{a \sin x - x \sin a}{ax^2 - xa^2}$

$\displaystyle = \lim \limits_{x \to a } \ \frac{a \sin x - x \sin a}{ax(x-a)}$

$\displaystyle \text{Let } t = x - a$

$\displaystyle \text{Then, as } x \rightarrow a, t \rightarrow 0$

$\displaystyle = \lim \limits_{t \to 0 } \ \frac{a \sin (t+a) - (t+a) \sin a}{a(t+a)t}$

$\displaystyle = \lim \limits_{t \to 0 } \ \frac{a \sin t \cos a + a \sin a \cos t - t \sin a - a \sin a}{a(t+a)t}$

$\displaystyle = \lim \limits_{t \to 0 } \ \frac{a \sin t \cos a + a \sin a (\cos t - 1) -t \sin a }{a(t+a)t}$

$\displaystyle = \lim \limits_{t \to 0 } \ \frac{a \sin t \cos a + a \sin a ( 2 \sin^2 \frac{t}{2} ) - t \sin a }{a(t+a)t}$

$\displaystyle = \lim \limits_{t \to 0 } \frac{a \sin t \cos a }{a ( t+a) t} + \lim \limits_{t \to 0 } \frac{a \sin a ( 2 \sin^2 \frac{t}{2})}{a(t+a)t} - \lim \limits_{t \to 0 } \frac{t \sin a}{a(t+a)t}$

$\displaystyle = \frac{a \cos a }{a^2}+ 0 - \frac{\sin a }{a^2}$

$\displaystyle = \frac{a \cos a - \sin a}{a^2}$

$\\$

$\displaystyle \text{ Question 10: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sqrt{2}- \sqrt{1+\sin x}}{\cos^2 x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sqrt{2}- \sqrt{1+\sin x}}{\cos^2 x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \Bigg( \frac{\sqrt{2}- \sqrt{1+\sin x}}{\cos^2 x} \times \frac{\sqrt{2}+ \sqrt{1+\sin x}}{\sqrt{2}+ \sqrt{1+\sin x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \frac{2 - 1 - \sin x}{\cos^2 x (\sqrt{2}+ \sqrt{1+\sin x}) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \frac{1 - \sin x}{(1 - \sin^2 x) (\sqrt{2}+ \sqrt{1+\sin x}) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \frac{1}{(1 + \sin x) (\sqrt{2}+ \sqrt{1+\sin x}) }$

$\displaystyle = \frac{1}{(1+1)(\sqrt{2}+\sqrt{2})}$

$\displaystyle = \frac{1}{4\sqrt{2}}$

$\\$

$\displaystyle \text{ Question 11: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sqrt{2 - \sin x}-1}{\Big( \frac{\pi}{2} - x \Big)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\sqrt{2 - \sin x}-1}{\Big( \frac{\pi}{2} - x \Big)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2 - \sin (\frac{\pi}{2} -h) }-1}{\Big( \frac{\pi}{2} - (\frac{\pi}{2}-h) \Big)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2 - \cos h }-1}{h^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{(\sqrt{2 - \cos h }-1)(\sqrt{2 - \cos h}+1)}{h^2(\sqrt{2 - \cos h}+1)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{2-\cos h - 1}{h^2(\sqrt{2 - \cos h}+1)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1-\cos h }{h^2(\sqrt{2 - \cos h}+1)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{h}{2} }{\frac{4h^2}{4} (\sqrt{2 - \cos h}+1)}$

$\displaystyle = \frac{1}{2} \lim \limits_{h \to 0 } \Bigg( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \Bigg)^2 \times \lim \limits_{h \to 0 } \Bigg( \frac{1}{\sqrt{2 - \cos h}+1} \Bigg)$

$\displaystyle = \frac{1}{2(\sqrt{2-1}+1)} = \frac{1}{4}$

$\\$

$\displaystyle \text{ Question 12: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2}- \cos x - \sin x}{\Big( \frac{\pi}{4} - x \Big)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2}- \cos x - \sin x}{\Big( \frac{\pi}{4} - x \Big)^2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1- \Big(\frac{\cos x}{\sqrt{2}} + \frac{\sin x}{\sqrt{2}} \Big)}{\frac{1}{\sqrt{2}}\Big( \frac{\pi}{4} - x \Big)^2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1- \Big( \cos \frac{\pi}{4} \cos x+ \sin \frac{\pi}{4} \sin x \Big)}{\frac{1}{\sqrt{2}}\Big( \frac{\pi}{4} - x \Big)^2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2} \Big(1- \cos (\frac{\pi}{4} - x ) \Big)}{\Big( \frac{\pi}{4} - x \Big)^2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2} \Bigg( 2 \sin^2 \Big( \frac{\frac{\pi}{4}-x}{2} \Big) \Bigg)}{\Big( \frac{\pi}{4} - x \Big)^2}$

$\displaystyle = 2 \sqrt{2}\lim \limits_{x \to \frac{\pi}{4} } \ \frac{ \Bigg( \sin^2 \Big( \frac{\frac{\pi}{4}-x}{2} \Big) \Bigg)}{4 \Big( \frac{\frac{\pi}{4} - x }{2}\Big)^2}$

$\displaystyle = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$

$\\$

$\displaystyle \text{ Question 13: } \lim \limits_{x \to \frac{\pi}{8} } \ \frac{\cot 4x - \cos 4x}{(\pi - 8 x)^3}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{8} } \ \frac{\cot 4x - \cos 4x}{(\pi - 8 x)^3}$

$\displaystyle \text{Let } h = x - \frac{\pi}{8}. \text{ when } x \rightarrow \frac{\pi}{8}, \text{ then } h \rightarrow 0$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\cot 4( \frac{\pi}{8} - h) - \cos 4( \frac{\pi}{8} - h)}{ \Big(\pi - 8 ( \frac{\pi}{8} - h) \Big)^3}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\tan 4h - \sin 4h}{(8h)^3}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sin 4h ( 1 - \cos 4h)}{(\cos 4h) 512h^3}$

$\displaystyle = \lim \limits_{h \to 0 } \Bigg( \frac{\tan 4h}{4h} \times \frac{2 \sin^2 2h}{32 \times 4h^2} \Bigg)$

$\displaystyle = \frac{1}{16} \lim \limits_{h \to 0 } \frac{\tan 4h}{4h} \times \lim \limits_{h \to 0 } \Bigg( \frac{\sin 2h}{2h} \Bigg)^2$

$\displaystyle = \frac{1}{16} \times 1 \times 1 = \frac{1}{16}$

$\\$

$\displaystyle \text{ Question 14: } \lim \limits_{x \to a } \ \frac{\cos x - \cos a}{\sqrt{x}-\sqrt{a}}$

Answer:

$\displaystyle \lim \limits_{x \to a } \ \frac{\cos x - \cos a}{\sqrt{x}-\sqrt{a}}$

$\displaystyle = \lim \limits_{x \to a } \ \frac{ -2 \sin (\frac{x+a}{2}) \sin(\frac{x-a}{2}) }{\sqrt{x}-\sqrt{a}}$

$\displaystyle = -2\lim \limits_{x \to a } \ \frac{ \sin (\frac{x+a}{2}) \sin(\frac{x-a}{2} ) (\sqrt{x}+\sqrt{a}) }{ 2(\frac{x-a}{2}) }$

$\displaystyle = -2\lim \limits_{x \to a } \sin \Big(\frac{x+a}{2} \Big) \times \frac{\sin (\frac{x-a}{2})}{2(\frac{x-a}{2})} (\sqrt{x}+\sqrt{a})$

$\displaystyle = - 2 \sin \Big(\frac{2a}{a} \Big) \Big( \frac{\sqrt{a}+\sqrt{a}}{2} \Big)$

$\displaystyle = -2 \sqrt{a} \ \sin a$

$\\$

$\displaystyle \text{ Question 15: } \lim \limits_{x \to \pi } \ \frac{\sqrt{5+\cos x}-2}{(\pi - x)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \pi } \ \frac{\sqrt{5+\cos x}-2}{(\pi - x)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{5+\cos (\pi - h)}-2}{(\pi - (\pi - h))^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{5 - \cos h}-2}{h^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{5 - \cos h}-2}{h^2} \times \frac{\sqrt{5 - \cos h}+2}{\sqrt{5 - \cos h}+2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{5 - \cos h - 4}{h^2(\sqrt{5 - \cos h}+ 2)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \cos h }{h^2(\sqrt{5 - \cos h}+ 2)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{h}{2} }{4(\frac{h^2}{4})(\sqrt{5 - \cos h}+ 2)}$

$\displaystyle = \frac{1}{2(\sqrt{5-1}+2)} = \frac{1}{2 \times 4} = \frac{1}{8}$

$\\$

$\displaystyle \text{ Question 16: } \lim \limits_{x \to a } \ \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x-a}$

Answer:

$\displaystyle \lim \limits_{x \to a } \ \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x-a}$

$\displaystyle = \lim \limits_{x \to a } \ \frac{-2 \sin (\frac{\sqrt{x}+\sqrt{a}}{2} ) \sin (\frac{\sqrt{x}-\sqrt{a}}{2})}{2(\frac{\sqrt{x}-\sqrt{a}}{2}) (\sqrt{x}+\sqrt{a})}$

$\displaystyle = \lim \limits_{x \to a } \Bigg( \frac{- \sin (\frac{\sqrt{x}+\sqrt{a}}{2} )}{\sqrt{x}+\sqrt{a}} \times \frac{\sin (\frac{\sqrt{x}-\sqrt{a}}{2} )}{\frac{\sqrt{x}-\sqrt{a}}{2}} \Bigg)$

$\displaystyle = \frac{-1 \sin \sqrt{a} }{2 \sqrt{a}} \times 1 = \frac{-\sin \sqrt{a}}{2\sqrt{a}}$

$\\$

$\displaystyle \text{ Question 17: } \lim \limits_{x \to a } \ \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x-a}$

Answer:

$\displaystyle \lim \limits_{x \to a } \ \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x-a}$

$\displaystyle = \lim \limits_{x \to a } \ \frac{2 \cos (\frac{\sqrt{x}+\sqrt{a}}{2}) \sin (\frac{\sqrt{x}-\sqrt{a}}{2})}{(\sqrt{x}+\sqrt{a})(\sqrt{x}-\sqrt{a})}$

$\displaystyle = \lim \limits_{x \to a } \ \frac{2 \cos (\frac{\sqrt{x}+\sqrt{a}}{2})}{\sqrt{x}+\sqrt{a}} \times \frac{\sin (\frac{\sqrt{x}-\sqrt{a}}{2})}{2\frac{\sqrt{x}-\sqrt{a}}{2}}$

$\displaystyle = \frac{1}{\sqrt{a}+\sqrt{a}} \cos \Big( \frac{2\sqrt{a}}{2} \Big) = \frac{1}{2\sqrt{a}} \cos \sqrt{a}$

$\\$

$\displaystyle \text{ Question 18: } \lim \limits_{x \to 1 } \ \frac{1-x^2}{\sin 2\pi x}$

Answer:

$\displaystyle \lim \limits_{x \to 1 } \ \frac{1-x^2}{\sin 2\pi x}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{1-(1-h)^2}{\sin 2\pi (1-h)}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{2h-h^2}{-\sin 2\pi h}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{-(2-h)}{\frac{\sin 2\pi h}{h}}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{(h-2)}{\frac{2\pi \sin 2\pi h}{2 \pi h}}$

$\displaystyle = \frac{0-2}{2\pi \times 1} = \frac{-1}{\pi}$

$\\$

$\displaystyle \text{ Question 19: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{f(x) - f(\frac{\pi}{4})}{x - \frac{\pi}{4}} , \text{ where} f(x) = \sin 2x$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{f(x) - f(\frac{\pi}{4})}{x - \frac{\pi}{4}}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{f(\frac{\pi}{4}+h) - f(\frac{\pi}{4})}{\frac{\pi}{4}+h - \frac{\pi}{4}}$

$\displaystyle \text{ Given } f(x) = \sin 2x$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\cos 2h - 1}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \ -2 \Bigg( \frac{\sin^2 h}{h \times h} \Bigg) \times h$

$\displaystyle = \lim \limits_{h \to 0 } (-2h) = 0$

$\\$

$\displaystyle \text{ Question 20: } \lim \limits_{x \to 1 } \ \frac{1 + \cos \pi x}{(1-x)^2 }$

Answer:

$\displaystyle \lim \limits_{x \to 1 } \ \frac{1 + \cos \pi x}{(1-x)^2 }$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{1 + \cos \pi (1-h)}{(1-(1-h))^2 }$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{1 - \cos \pi h }{h^2 }$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{\pi h}{2} }{\frac{4}{\pi^2}(\frac{\pi^2h^2}{4}) }$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{\sin^2 \frac{\pi h}{2} }{\frac{2}{\pi^2}(\frac{\pi^2h^2}{4}) }$

$\displaystyle = \frac{\pi^2}{2}$

$\\$

$\displaystyle \text{ Question 21: } \lim \limits_{x \to 1 } \ \frac{1-x^2}{\sin \pi x}$

Answer:

$\displaystyle \lim \limits_{x \to 1 } \ \frac{1-x^2}{\sin \pi x}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{1-(1-h)^2}{\sin \pi (1-h)}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{h(2-h)}{\sin \pi h}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{(2-h)}{\pi \times \frac{\sin \pi h}{\pi h}}$

$\displaystyle = \frac{2-0}{\pi} = \frac{2}{\pi}$

$\\$

$\displaystyle \text{ Question 22: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \sin 2x}{1 + \cos 4x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \sin 2x}{1 + \cos 4x}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \sin 2(\frac{\pi}{4} - h)}{1 + \cos 4(\frac{\pi}{4} - h)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \cos 2h }{1 - \cos 4h}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{2 \sin^2 h }{2 \sin^2 2h}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\frac{ \sin^2 h}{h^2} }{\frac{4 \sin^2 2h}{4h^2} }$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{ Question 23: } \lim \limits_{x \to \pi } \ \frac{1+\cos x }{\tan^2 x}$

Answer:

$\displaystyle \lim \limits_{x \to \pi } \ \frac{1+\cos x }{\tan^2 x}$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{\sin^2 x} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{1 - \cos^2 x} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{(1 - \cos x)(1+ \cos x)} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{\cos^2 x }{(1 - \cos x)}$

$\displaystyle = \frac{\cos^2 \pi }{(1 - \cos \pi)} = \frac{(-1)^2}{1-(-1)} = \frac{1}{2}$

$\\$

$\displaystyle \text{ Question 24: } \lim \limits_{n \to \infty } \ n \sin \Big( \frac{\pi}{4n} \Big) \cos \Big( \frac{\pi}{4n} \Big)$

Answer:

$\displaystyle \lim \limits_{n \to \infty } \ n \sin \Big( \frac{\pi}{4n} \Big) \cos \Big( \frac{\pi}{4n} \Big)$

$\displaystyle = \lim \limits_{n \to \infty } \ n \sin \Big( \frac{\pi}{4n} \Big) \times \lim \limits_{n \to \infty } \cos \Big( \frac{\pi}{4n} \Big)$

$\displaystyle = \lim \limits_{n \to \infty } \ \frac{ \sin \Big( \frac{\pi}{4n} \Big)}{\frac{\pi}{4n}} \times \frac{\pi}{4} \times 1$

$\displaystyle = \frac{\pi}{4}$

$\\$

$\displaystyle \text{ Question 25: } \lim \limits_{n \to \infty } \ 2^{n-1} \sin \Big( \frac{a}{2^n} \Big)$

Answer:

$\displaystyle \lim \limits_{n \to \infty } \ 2^{n-1} \sin \Big( \frac{a}{2^n} \Big)$

$\displaystyle = \lim \limits_{n \to \infty } \frac{2^n}{2} \times \frac{\sin (\frac{a}{2^n})}{ (\frac{a}{2^n}) } \times (\frac{a}{2^n})$

$\displaystyle = \frac{a}{2}$

$\\$

$\displaystyle \text{ Question 26: } \lim \limits_{n \to \infty } \ \frac{\sin \Big( \frac{a}{2^n} \Big)}{\sin \Big( \frac{b}{2^n} \Big)}$

Answer:

$\displaystyle \lim \limits_{n \to \infty } \ \frac{\sin \Big( \frac{a}{2^n} \Big)}{\sin \Big( \frac{b}{2^n} \Big)}$

$\displaystyle = \lim \limits_{n \to \infty } \frac{(\frac{a}{2^n}) \sin (\frac{a}{2^n}) }{(\frac{a}{2^n}) \times (\frac{b}{2^n}) \times \frac{\sin \frac{b}{2^n}}{\frac{b}{2^n}} }$

$\displaystyle = \frac{a}{b}$

$\\$

$\displaystyle \text{ Question 27: } \lim \limits_{x \to -1 } \ \frac{x^2-x-2}{(x^2+x) + \sin ( x+1)}$

Answer:

$\displaystyle \lim \limits_{x \to -1 } \ \frac{x^2-x-2}{(x^2+x) + \sin ( x+1)}$

$\displaystyle = \lim \limits_{x \to -1 } \ \frac{(x-2)(x+1)}{x(x+1) + \sin ( x+1)}$

$\displaystyle = \lim \limits_{x \to -1 } \ \frac{1}{\frac{x(x+1)}{(x-2)(x+1)} + \frac{\sin (x+1)}{(x-2)(x+1)} }$

$\displaystyle = \lim \limits_{x \to -1 } \ \frac{1}{\frac{x}{(x-2)} + \frac{\sin (x+1)}{(x-2)(x+1)} }$

$\displaystyle = \lim \limits_{x \to -1 } \ \frac{x-2}{\frac{x}{1} + \frac{\sin (x+1)}{(x+1)} }$

$\displaystyle = \frac{-1-2}{-1+1} = \frac{-3}{0} = -\infty$

$\\$

$\displaystyle \text{ Question 28: } \lim \limits_{x \to 2 } \ \frac{x^2 - x - 2}{x^2 - 2x + \sin (x-2)}$

Answer:

$\displaystyle \lim \limits_{x \to 2 } \ \frac{x^2 - x - 2}{x^2 - 2x + \sin (x-2)}$

$\displaystyle = \lim \limits_{x \to 2 } \ \frac{(x-2)(x+1)}{x(x-2) + \sin (x-2)}$

$\displaystyle = \lim \limits_{x \to 2 } \ \frac{1}{\frac{x}{x+1}+ \frac{\sin (x-2)}{(x-2)(x+1)} }$

$\displaystyle = \lim \limits_{x \to 2 } \ \frac{x+1}{\frac{x}{1}+ \frac{\sin (x-2)}{(x-2)} }$

$\displaystyle = \frac{2+1}{2+1} = 1$

$\\$

$\displaystyle \text{ Question 29: } \lim \limits_{x \to 1 } \ (1-x) \tan \Big( \frac{\pi x}{2} \Big)$

Answer:

$\displaystyle \lim \limits_{x \to 1 } \ (1-x) \tan \Big( \frac{\pi x}{2} \Big)$

$\displaystyle = \lim \limits_{h \to 0 } \ [1-(1-h)] \tan \Big( \frac{\pi (1-h)}{2} \Big)$

$\displaystyle = \lim \limits_{h \to 0 } \ h \cot \frac{\pi h}{2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{h}{\tan \frac{\pi h}{2} }$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1}{ \frac{\tan \frac{\pi h}{2} \times \frac{\pi }{2}}{\frac{\pi h}{2}} }$

$\displaystyle = \frac{2}{\pi}$

$\\$

$\displaystyle \text{ Question 30: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \tan x}{1 - \sqrt{2} \sin x}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1 - \tan x}{1 - \sqrt{2} \sin x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(1 - \tan x)(1 + \sqrt{2} \sin x)}{(1 - \sqrt{2} \sin x)(1 + \sqrt{2} \sin x)}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(1 - \tan x)(1 + \sqrt{2} \sin x)}{1 - 2 \sin^2 x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(1 - \frac{\sin x}{\cos x} )(1 + \sqrt{2} \sin x)}{\cos 2x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(\cos x - \sin x )(1 + \sqrt{2} \sin x)}{\cos x \cos 2x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(\cos x - \sin x )(1 + \sqrt{2} \sin x)}{\cos x (\cos^2 x - \sin^2 x)}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(\cos x - \sin x )(1 + \sqrt{2} \sin x)}{\cos x (\cos x - \sin x)(\cos x + \sin x)}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(1 + \sqrt{2} \sin x)}{\cos x (\cos x + \sin x)}$

$\displaystyle = \frac{1 + \sqrt{2} \times \frac{1}{\sqrt{2}}}{(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}) } = \frac{2}{\frac{1}{\sqrt{2}} \times \sqrt{2}} = 2$

$\\$

$\displaystyle \text{ Question 31: } \lim \limits_{x \to \pi } \ \frac{\sqrt{2+\cos x}-1}{(\pi - x)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \pi } \ \frac{\sqrt{2+\cos x}-1}{(\pi - x)^2}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{\sqrt{2+\cos (\pi - h)}-1}{(\pi - (\pi - h))^2}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{\sqrt{2-\cos h}-1}{h^2}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{\sqrt{2-\cos h}-1}{h^2} \times \frac{\sqrt{2-\cos h}+1}{\sqrt{2-\cos h}+1}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{2 - \cos h - 1}{h^2(\sqrt{2-\cos h}+1)}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{1 - \cos h }{h^2(\sqrt{2-\cos h}+1)}$

$\displaystyle =\lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{h}{2} }{4 \times \frac{h^2}{4} \times (\sqrt{2-\cos h}+1)}$

$\displaystyle = \frac{1 \times 1^2}{2(\sqrt{2-1} + 1 )} = \frac{1}{4}$

$\\$

$\displaystyle \text{ Question 32: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \Bigg( \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}} \Bigg) \Bigg( \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \frac{\cos x - \sin x}{(x - \frac{\pi}{4}) ( \sqrt{\cos x} + \sqrt{\sin x} ) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \frac{\cos (\frac{\pi}{4}+h) - \sin (\frac{\pi}{4}+h)}{(\frac{\pi}{4}+h - \frac{\pi}{4}) \Big( \sqrt{\cos (\frac{\pi}{4}+h)} + \sqrt{\sin (\frac{\pi}{4}+h)} \Big) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \frac{ \cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h - \sin\frac{\pi}{4} \cos h - \cos \frac{\pi}{4} \sin h }{(h) \Big( \sqrt{\cos (\frac{\pi}{4}+h)} + \sqrt{\sin (\frac{\pi}{4}+h)} \Big) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \frac{-\sqrt{2} \sin h }{(h) \Big( \sqrt{\cos (\frac{\pi}{4}+h)} + \sqrt{\sin (\frac{\pi}{4}+h)} \Big) }$

$\displaystyle = \frac{-\sqrt{2} \times 1}{2 ( \frac{1}{2})^{\frac{1}{4}} } = \frac{-1}{2^{\frac{1}{4}}} = -2^{-\frac{1}{4} }$

$\\$

$\displaystyle \text{ Question 33: } \lim \limits_{x \to 1 } \ \frac{1 - \frac{1}{x} }{\sin \pi ( x-1)}$

Answer:

$\displaystyle \lim \limits_{x \to 1 } \ \frac{1 - \frac{1}{x} }{\sin \pi ( x-1)}$

$\displaystyle = \lim \limits_{x \to 1 } \ \frac{x-1}{x \sin \pi ( x-1)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{h+1-1}{(h+1) \sin \pi (h+1-1)}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{h}{(h+1) \sin \pi h}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1}{\pi (h+1) \frac{\sin \pi h}{\pi h} }$

$\displaystyle = \frac{1}{\pi(0+1) \times 1} = \frac{1}{\pi}$

$\\$

$\displaystyle \text{ Question 34: } \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\cot^2 x - 3}{\mathrm{cosec} - 2}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\cot^2 x - 3}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\mathrm{cosec}^2 x - 1 - 3}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\mathrm{cosec}^2 x - 4}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{(\mathrm{cosec} x - 2)(\mathrm{cosec} x + 2)}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \mathrm{cosec} x + 2$

$\displaystyle = 4$

$\\$

$\displaystyle \text{ Question 35: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2}- \cos x - \sin x }{(4x-\pi)^2}$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2}- \cos x - \sin x }{(4x-\pi)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2}- \{ \cos (\frac{\pi}{4}+h) + \sin (\frac{\pi}{4}+h) \} }{(4(\frac{\pi}{4}+h)-\pi)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2}- \{ \cos (\frac{\pi}{4}+h) + \sin (\frac{\pi}{4}+h) \} }{(4h)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2}- \{ \cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h + \sin \frac{\pi}{4} \cos h + \cos \frac{\pi}{4} \sin h \} }{(4h)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2}- \sqrt{2} \cos h}{(4h)^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sqrt{2}(1- \cos h)}{16h^2}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{ 2 \sqrt{2} \sin^2 \frac{h}{2} }{64 (\frac{h^2}{4}) }$

$\displaystyle = \frac{\sqrt{2}}{32} \times 1^2 = \frac{1}{16\sqrt{2}}$

$\\$

$\displaystyle \text{ Question 36: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\Big( \frac{\pi}{2} -x \Big) \sin x - 2 \cos x}{\Big( \frac{\pi}{2} -x \Big)+ \cot x }$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{\Big( \frac{\pi}{2} -x \Big) \sin x - 2 \cos x}{\Big( \frac{\pi}{2} -x \Big)+ \cot x }$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\Big( \frac{\pi}{2} -(\frac{\pi}{2} - h) \Big) \sin (\frac{\pi}{2} - h) - 2 \cos (\frac{\pi}{2} - h)}{\Big( \frac{\pi}{2} -(\frac{\pi}{2} - h) \Big)+ \cot (\frac{\pi}{2} - h) }$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{h \cos h - 2 \sin h }{h + \tan h}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{ \cos h - 2 \frac{\sin h}{h} }{1 + \frac{\tan h}{h} }$

$\displaystyle = \frac{1 - 2 (1)}{1+1} = \frac{-1}{2}$

$\\$

$\displaystyle \text{ Question 37: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\cos x - \sin x }{\Big( \frac{\pi}{4} - x \Big) (\cos x + \sin x ) }$

Answer:

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\cos x - \sin x }{\Big( \frac{\pi}{4} - x \Big) (\cos x + \sin x ) }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}}\sin x }{\Big( \frac{\pi}{4} - x \Big) \frac{(\cos x + \sin x )}{\sqrt{2}} }$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\sqrt{2} ( \sin \frac{\pi}{4} \cos x - \cos \frac{\pi}{4} \sin x ) }{(\frac{\pi}{4} - x) (\cos x + \sin x)}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \Bigg( \frac{\sqrt{2}}{\cos x + \sin x} \times \frac{\sin (\frac{\pi}{4} - x) }{\frac{\pi}{4} - x } \Bigg)$

$\displaystyle = \frac{\sqrt{2}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} } \times 1 = 1$

$\\$

$\displaystyle \text{ Question 38: } \lim \limits_{x \to \pi } \ \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} \Big( \cos \frac{x}{4} - \sin \frac{x}{4} \Big) }$

Answer:

$\displaystyle \lim \limits_{x \to \pi } \ \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} \Big( \cos \frac{x}{4} - \sin \frac{x}{4} \Big) }$

$\displaystyle \text{Put } x = \pi + h, \text{ then when } x \rightarrow \pi, h \rightarrow 0$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \sin \frac{\pi + h}{2}}{\cos \frac{\pi + h}{2} \Big( \cos \frac{\pi + h}{4} - \sin \frac{\pi + h}{4} \Big) }$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \sin ( \frac{\pi}{2} + \frac{h}{2} ) }{\cos ( \frac{\pi}{2} + \frac{h}{2} ) [ \cos ( \frac{\pi}{2} + \frac{h}{2} ) - \sin ( \frac{\pi}{2} + \frac{h}{2} ) ]}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \cos \frac{h}{2} }{-\sin \frac{h}{2} [ \cos ( \frac{\pi}{2} + \frac{h}{2} ) - \sin ( \frac{\pi}{2} + \frac{h}{2} ) ]}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{1 - \cos \frac{h}{2} }{-\sin \frac{h}{2} [ ( \cos \frac{\pi}{4} \cos \frac{h}{4} - \sin \frac{\pi}{4} \sin \frac{h}{4} ) - ( \sin \frac{\pi}{4} \cos \frac{h}{4} + \cos \frac{\pi}{4} \sin \frac{h}{4} ) ]}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{2 \sin^2 \frac{h}{4} }{-2\sin \frac{h}{4} \cos \frac{h}{4} [ ( \frac{1}{\sqrt{2}} \cos \frac{h}{4} - \frac{1}{\sqrt{2}} \sin \frac{h}{4} ) - ( \frac{1}{\sqrt{2}} \cos \frac{h}{4} + \frac{1}{\sqrt{2}} \sin \frac{h}{4} ) ]}$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{\sin \frac{h}{4} }{-\cos \frac{h}{4} \times ( - \sqrt{2} \sin \frac{h}{4} )}$

$\displaystyle = \frac{1}{ \sqrt{2} } \times 1 = \frac{1}{\sqrt{2}}$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Limits – Exercise 29.8

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