Class 11: Limits – Exercise 29.7 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{ \sin x}{ x} = 1$ $\displaystyle \lim \limits_{x \to 0 } \cos x = 1$

$\displaystyle \lim \limits_{x \to 0 } \ \frac{ \tan x}{ x} = 1$

Evaluate the following limits:

$\displaystyle \text{ Question 1: } \lim \limits_{x \to 0 } \ \frac{ \sin 3x}{ 5x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{ \sin 3x}{ 5x} = \frac{1}{5} \lim \limits_{x \to 0 } \ \frac{ \sin 3x}{ 3x} \times 3 = \frac{1}{5} \times 1 \times 3 = \frac{3}{5}$

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$\displaystyle \text{ Question 2: } \lim \limits_{x \to 0 } \ \frac{\sin x^0}{ x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin x^0}{ x} = \lim \limits_{x \to 0 } \ \frac{ \sin \frac{\pi}{180} x}{ \frac{\pi}{180} x} \times \frac{\pi}{180} = \frac{\pi}{180} \times 1 = \frac{\pi}{180}$

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$\displaystyle \text{ Question 3: } \lim \limits_{x \to 0 } \ \frac{x^2}{\sin x^2}$

Answer:

$\displaystyle \text{Given: } \lim \limits_{x \to 0 } \ \frac{x^2}{\sin x^2}$

$\displaystyle \text{When } x \rightarrow 0, \text{ then } x^2 \rightarrow 0$

$\displaystyle \text{Let } \theta = x^2$

$\displaystyle \Rightarrow \lim \limits_{x \to 0 } \ \frac{\theta}{\sin \theta} = \lim \limits_{x \to 0 } \ \frac{1}{\frac{\sin \theta}{\theta}} = 1$

$\\$

$\displaystyle \text{ Question 4: } \lim \limits_{x \to 0 } \ \frac{\sin x \cos x}{3x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin x \cos x}{3x} = \frac{1}{3} \lim \limits_{x \to 0 } \ \frac{\sin x}{x} \times \cos x = \frac{1}{3} \times 1 = \frac{1}{3}$

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$\displaystyle \text{ Question 5: } \lim \limits_{x \to 0 } \ \frac{3 \sin x - 4 \sin^3 x}{x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{3 \sin x - 4 \sin^3 x}{x} = \lim \limits_{x \to 0 } \ \frac{ \sin 3x}{x} = \lim \limits_{x \to 0 } \ \frac{ \sin 3x}{3x} \times 3 = 1 \times 3 = 3$

$\displaystyle \text{Note: } \sin 3x = 3 \sin x - 4 \sin^3 x$

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$\displaystyle \text{ Question 6: } \lim \limits_{x \to 0 } \ \frac{\tan 8x}{\sin 2x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\tan 8x}{\sin 2x} = \lim \limits_{x \to 0 } \ \frac{\tan 8x}{8x} \times \frac{8x}{\frac{\sin 2x}{2x} \times 2x} = 4$

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$\displaystyle \text{ Question 7: } \lim \limits_{x \to 0 } \ \frac{\tan mx}{\tan nx}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\tan mx}{\tan nx} = \lim \limits_{x \to 0 } \ \frac{\tan mx}{mx} \times \frac{mx}{\frac{\tan nx}{x}} \times nx = \frac{m}{n}$

$\\$

$\displaystyle \text{ Question 8: } \lim \limits_{x \to 0 } \ \frac{\sin 5x}{\tan 3x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin 5x}{\tan 3x} = \lim \limits_{x \to 0 } \ \frac{\sin 5x}{5x} \times \frac{5x}{\frac{\tan 3x}{3x} \times 3x } = \frac{5}{3}$

$\\$

$\displaystyle \text{ Question 9: } \lim \limits_{x \to 0 } \ \frac{\sin x^{\circ}}{x^{\circ}}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin x^{\circ}}{x^{\circ}} \text{ This is of the form } \frac{\infty}{\infty}$

$\displaystyle \text{Let } y = x^n$

$\displaystyle \text{Therefore } \lim \limits_{x \to 0 } \ \frac{\sin x^{\circ}}{x^{\circ}} = \lim \limits_{y \to 0 } \ \frac{\sin y}{y} = 1$

$\\$

$\displaystyle \text{ Question 10: } \lim \limits_{x \to 0 } \ \frac{7x \cos x - 3 \sin x}{4x + \tan x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{7x \cos x - 3 \sin x}{4x + \tan x} \text{ This is of the form } \frac{\infty}{\infty}$

Dividing the numerator and denominator by $x$

$\displaystyle \lim \limits_{x \to 0 } \ \frac{7x \cos x - 3 \sin x}{4x + \tan x} = \lim \limits_{x \to 0 } \ \frac{7 \cos x - 3 \frac{\sin x}{x}}{4 + \frac{\tan x}{x}} = \frac{7 \times 1 - 3 \times 1}{4+1} = \frac{4}{5}$

$\\$

$\displaystyle \text{ Question 11: } \lim \limits_{x \to 0 } \ \frac{\cos ax - \cos bx}{\cos cx - \cos dx}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\cos ax - \cos bx}{\cos cx - \cos dx} \text{ This is of the form } \frac{\infty}{\infty}$

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\cos ax - \cos bx}{\cos cx - \cos dx}$

$\displaystyle = \lim \limits_{x \to 0 } \ \frac{-2\sin(\frac{ax-bx}{2}) \sin(\frac{ax+bx}{2})}{-2 \sin(\frac{cx+dx}{2}) \sin(\frac{cx-dx}{2}) }$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin (\frac{ax+bx}{2} )}{\frac{ax+bx}{2}} \times (\frac{ax+bx}{2}) \times \frac{\sin (\frac{ax-bx}{2} )}{\frac{ax+bx}{2}} \times (\frac{ax-bx}{2}) }{\frac{\sin (\frac{cx+dx}{2} )}{\frac{cx+dx}{2}} \times (\frac{cx+dx}{2}) \times \frac{\sin (\frac{cx-dx}{2} )}{\frac{cx+dx}{2}} \times (\frac{cx-dx}{2}) } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ (\frac{ax+bx}{2})(\frac{ax-bx}{2}) }{ (\frac{cx+dx}{2})(\frac{cx-dx}{2}) } \Bigg]$

$\displaystyle = \frac{x^2(a^2-b^2)}{x^2(c^2-d^2)}$

$\displaystyle = \frac{a^2-b^2}{c^2-d^2}$

$\\$

$\displaystyle \text{ Question 12: } \lim \limits_{x \to 0 } \ \frac{\tan^2 3x}{x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\tan^2 3x}{x^2} = \lim \limits_{x \to 0 } \ \frac{\tan 3x}{3x} \times \frac{\tan 3x}{3x} \times 9 = 1 \times 1 \times 9 = 9$

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$\displaystyle \text{ Question 13: } \lim \limits_{x \to 0 } \ \frac{1 - \cos mx}{x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{1 - \cos mx}{x^2} = \lim \limits_{x \to 0 } \ \frac{2 \sin^2 \frac{mx}{2}}{x^2} =2 \lim \limits_{x \to 0 } \ \frac{\sin \frac{mx}{2}}{\frac{mx}{2}} \times \frac{\sin \frac{mx}{2}}{\frac{mx}{2}} \times \frac{m}{2} \times \frac{m}{2} \\ \\ {\hspace{6.0cm} = 2 \times \frac{m}{2} \times \frac{m}{2} = \frac{m^2}{2}}$

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$\displaystyle \text{ Question 14: } \lim \limits_{x \to 0 } \ \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x} = \lim \limits_{x \to 0 } \ \frac{ \frac{3 \sin 2x}{x} + 2}{3 + \frac{2 \tan 3x}{x}} = \lim \limits_{x \to 0 } \ \frac{ \frac{3 \sin 2x}{2x} \times 2+ 2}{3 + \frac{2 \tan 3x}{3x} \times 3} = \frac{6+2}{3+6} = \frac{8}{9}$

$\\$

$\displaystyle \text{ Question 15: } \lim \limits_{x \to 0 } \ \frac{\cos 3x - \cos 7x}{x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\cos 3x - \cos 7x}{x^2} \\ \\ = \lim \limits_{x \to 0 } \ \frac{ -2\sin (\frac{3x+7x}{2}) \sin(\frac{3x-7x}{2}) }{ x^2 } \\ \\ = \lim \limits_{x \to 0 } \ \frac{-2\sin 5x \sin (-2x)}{x^2} \\ \\ = \lim \limits_{x \to 0 } \ \frac{-2\sin 5x \sin 2x}{x^2} \\ \\ = \lim \limits_{x \to 0 } \ \frac{\sin 5x}{5x} \times \frac{\sin 2x}{2x} \times 5 \times 2 = 2 \times 5 \times 2 = 20$

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$\displaystyle \text{ Question 16: } \lim \limits_{\theta \to 0 } \ \frac{\sin 3 \theta}{\tan 2 \theta}$

Answer:

$\displaystyle \lim \limits_{\theta \to 0 } \ \frac{\sin 3 \theta}{\tan 2 \theta} = \lim \limits_{\theta \to 0 } \ \frac{\sin \theta}{3\theta} \times \frac{3\theta}{\frac{\tan 2 \theta}{2\theta} \times 2 \theta} = \frac{3}{2}$

$\\$

$\displaystyle \text{ Question 17: } \lim \limits_{x \to 0 } \ \frac{\sin x^2 ( 1- \cos x^2)}{x^6}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin x^2 ( 1- \cos x^2)}{x^6} \\ \\ = \lim \limits_{x \to 0 } \ \frac{\sin x^2 2 \sin^2 ( \frac{x^2}{2})}{x^6} \\ \\ = 2 \lim \limits_{x \to 0 } \ \frac{\sin x^2}{x^2} \times \frac{\sin \frac{x^2}{2}}{2 \times \frac{x^2}{2}} \times \frac{\sin \frac{x^2}{2} }{2 \times \frac{x^2}{2} } \\ \\ = \frac{2}{2\times2} = \frac{1}{2}$

$\\$

$\displaystyle \text{ Question 18: } \lim \limits_{x \to 0 } \ \frac{\sin^2 4x^2}{x^4}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin^2 4x^2}{x^4} = \lim \limits_{x \to 0 } \ \frac{\sin 4x^2}{x^2} \times \frac{\sin 4x^2}{x^2} = \lim \limits_{x \to 0 } \ \frac{\sin 4x^2}{4x^2} \times 4 \times \frac{\sin 4x^2}{4x^2} \times 4 \\ \\ { \hspace{7.0cm} = 4 \times 4 = 16}$

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$\displaystyle \text{ Question 19: } \lim \limits_{x \to 0 } \ \frac{x \cos x + 2 \sin x}{x^2 + \tan x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{x \cos x + 2 \sin x}{x^2 + \tan x} = \lim \limits_{x \to 0 } \ \frac{\cos x + \frac{2\sin x}{x}}{x + \frac{\tan x}{x} } = \frac{1+2\times 1}{0+1} = 3$

$\\$

$\displaystyle \text{ Question 20: } \lim \limits_{x \to 0 } \ \frac{2x - \sin x}{\tan x + x}$

Answer:

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

$\\$

$\displaystyle \text{ Question 21: } \lim \limits_{x \to 0 } \ \frac{5x \cos x + 3 \sin x}{3x^2 + \tan x }$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{5x \cos x + 3 \sin x}{3x^2 + \tan x } = \lim \limits_{x \to 0 } \ \frac{5 \cos x + 3 \Big(\frac{\sin x}{x} \Big) }{3x + \frac{\tan x}{x} } = \frac{5\cos 0+3}{3\times 0 + 1} = 8$

$\\$

$\displaystyle \text{ Question 22: } \lim \limits_{x \to 0 } \ \frac{\sin 3x - \sin x}{\sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin 3x - \sin x}{\sin x} = \lim \limits_{x \to 0 } \ \frac{2 \cos \frac{3x+x}{2} \sin \frac{3x-x}{2} }{\sin x} = \lim \limits_{x \to 0 } \ \frac{2 \cos 2x \cdot \sin x}{\sin x} = 2$

$\\$

$\displaystyle \text{ Question 23: } \lim \limits_{x \to 0 } \ \frac{\sin 5x - \sin 3x}{\sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin 5x - \sin 3x}{\sin x} = \lim \limits_{x \to 0 } \ \frac{2 \cos \frac{5x+3x}{2} \sin \frac{5x-3x}{2} }{\sin x} = \lim \limits_{x \to 0 } \ \frac{2 \cos 4x \cdot \sin x}{\sin x} = 2$

$\\$

$\displaystyle \text{ Question 24: } \lim \limits_{x \to 0 } \ \frac{\cos 3x - \cos 5x}{x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\cos 3x - \cos 5x}{x^2} \\ \\ = \lim \limits_{x \to 0 } \ \frac{-2 \sin \frac{3x+5x}{2} \sin \frac{3x-5x}{2} }{x^2} \\ \\ = 2 \lim \limits_{x \to 0 } \ \frac{ 2\sin 4x \cdot \sin x}{x^2} \\ \\ = \lim \limits_{x \to 0 } \ \frac{\sin 4x}{4x} \times 4 \times \frac{\sin x }{x} = 2 \times 4 = 8$

$\\$

$\displaystyle \text{ Question 25: } \lim \limits_{x \to 0 } \ \frac{\tan 3x - 2x}{3x - \sin^2 x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\tan 3x - 2x}{3x - \sin^2 x} = \lim \limits_{x \to 0 } \ \frac{\frac{\tan 3x}{x} - 2}{3 - \frac{\sin^2 x}{x}} = \lim \limits_{x \to 0 } \ \frac{\frac{\tan 3x}{3x}\times 3 - 2}{3 - \frac{\sin x}{x} \times \sin x} = \frac{3-2}{3-1\times 0} = \frac{1}{3}$

$\\$

$\displaystyle \text{ Question 26: } \lim \limits_{x \to 0 } \ \frac{\sin (2+x) - \sin(2-x)}{x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin (2+x) - \sin(2-x)}{x} = \lim \limits_{x \to 0 } \ \frac{2 \cos \frac{2+x+2-x}{2} \sin \frac{2+x-2+x}{2} }{x} = \lim \limits_{x \to 0 } \ \frac{2 \cos 2 \sin x}{x} \\ \\ {\hspace{11.0cm} = 2 \cos 2 }$

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$\displaystyle \text{ Question 27: } \lim \limits_{h \to 0 } \ \frac{(a+h)^2 \sin (a+h) - a^2 \sin a}{h}$

Answer:

$\displaystyle \lim \limits_{h \to 0 } \ \Bigg[ \frac{(a+h)^2 \sin (a+h) - a^2 \sin a}{h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{ a^2 \sin(a+h) + h^2 \sin (a+h) + 2ah \sin ( a+h) - a^2 \sin a }{h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \Bigg[ \ a^2 \Bigg\{ \frac{\sin(a+h) - \sin a}{h} \Bigg\} + \frac{h^2 \sin (a+h)}{h} + \frac{2ah \sin (a+h)}{h} \Bigg]$

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

$\displaystyle = \lim \limits_{h \to 0 } \Bigg[ \ a^2 \cos (\frac{a+h+a}{2} ) + h \sin (a+h) + 2a \sin ( a+h) \Bigg]$

$\displaystyle = a^2 \cos a + 2a \sin a$

$\\$

$\displaystyle \text{ Question 28: } \lim \limits_{x \to 0 } \ \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}$

Answer:

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

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin x}{\cos x} - \sin x}{3 \sin x - 4 \sin^3 x - 3 \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin x ( 1 - \cos x)}{\cos x ( -4 \sin^3 x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \sin^2 \frac{x}{2} }{\cos x ( -4 \sin^2 x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \sin \frac{x}{2} \times \sin \frac{x}{2} }{\cos x ( -4 \sin x) \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \frac{\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{x}{2} \times \frac{\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{x}{2} }{\cos x ( -4 \frac{\sin x}{x} \times x) (\frac{\sin x}{x} \times x) } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \times \frac{x}{2} \times \frac{x}{2} }{\cos x \times (-4) \times x \times x } \Bigg]$

$\displaystyle = \frac{-1}{8 \cos 0} = -\frac{1}{8}$

$\\$

$\displaystyle \text{ Question 29: } \lim \limits_{x \to 0 } \ \frac{\sec 5x- \sec 3x}{\sec3 x-\sec x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sec 5x- \sec 3x}{\sec3 x-\sec x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\frac{1}{\cos 5x}- \frac{1}{\cos 3x} }{\frac{1}{\cos 3x} -\frac{1}{\cos x}} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\cos 3x - \cos 5x}{\cos 5x \cos 3x \Big\{ \frac{\cos x - \cos 3x}{\cos x \cos 3x} \Big\}} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \Big(-2 \sin (\frac{3x+5x}{2}) \sin (\frac{3x-5x}{2}) \Big) \cos x}{\cos 5x \Big(-2 \sin (\frac{x+3x}{2}) \sin (\frac{x-3x}{2}) \Big)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 4x \times \sin (-x) \times \cos x}{\cos 5x \times \sin 2x \times \sin (-x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 4x}{4x} \times \frac{4x}{\frac{\sin 2x}{2x} \times 2x} \times \frac{\cos x}{\cos 5x} \Bigg]$

$\displaystyle = \frac{4 \cos 0}{2 \cos 0} = 2$

$\\$

$\displaystyle \text{ Question 30: } \lim \limits_{x \to 0 } \ \frac{1 - \cos 2x}{\cos 2x - \cos 8x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{1 - \cos 2x}{\cos 2x - \cos 8x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 x}{2 \sin (\frac{2x+8x}{2}) \sin (\frac{2x-8x}{2})} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin x \cdot \sin x}{2 \sin 5x \cdot \sin 3x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin x \cdot \sin x}{\sin 5x \cdot \sin 3x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\frac{\sin x}{x} \times x \times \frac{ \sin x}{x} \times x}{\frac{\sin 5x}{5x} \times 5x \times \frac{\sin 3x}{3x} \times 3x} \Bigg]$

$\displaystyle = \frac{1}{15}$

$\\$

$\displaystyle \text{ Question 31: } \lim \limits_{x \to 0 } \ \frac{1- \cos 2x + \tan^2 x}{x\sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{1- \cos 2x + \tan^2 x}{x\sin x} \\ \\ \\ = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 x + \tan^2 x}{x \sin x} \Bigg] \\ \\ \\ = \lim \limits_{x \to 0 } \ \frac{\frac{2\sin^2 x}{x^2} + \frac{\tan^2 x}{x^2}}{\frac{\sin x}{x}} \\ \\ \\ = \frac{2(1)^2 + (1)^2}{1} = 3$

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$\displaystyle \text{ Question 32: } \lim \limits_{x \to 0 } \ \frac{\sin ( a+ x) + \sin ( a-x) - 2 \sin a}{x \sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin ( a+ x) + \sin ( a-x) - 2 \sin a}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin (\frac{a+x+a-x}{2}) \cos (\frac{a+x-a+x}{2}) - 2 \sin a}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \sin a \cos x- 2 \sin a}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin a ( \cos x - 1)}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin a ( 1-2 \sin^2 \frac{x}{2} - 1)}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin a ( -2 \sin^2 \frac{x}{2})}{x \sin x} \Bigg]$

$\displaystyle = - 4 \sin a \lim \limits_{x \to 0 } \ \Bigg[ \frac{1}{ \frac{x \sin x}{x^2}} \times \frac{ \sin \frac{x}{2} }{ \frac{x}{2} } \times \frac{ \sin \frac{x}{2} }{ \frac{x}{2} } \times \frac{1}{4} \Bigg]$

$\displaystyle = - 4 \sin a \times \frac{1}{1} \times 1 \times 1 \times\frac{1}{4}$

$\displaystyle = -\sin a$

$\\$

$\displaystyle \text{ Question 33: } \lim \limits_{x \to 0 } \ \frac{x^2 - \tan 2x }{\tan x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{x^2 - \tan 2x }{\tan x} \Bigg] = \lim \limits_{x \to 0 } \Bigg[ \frac{x - \frac{\tan 2x}{x}}{\frac{\tan x}{x}} \Bigg] = \lim \limits_{x \to 0 } \Bigg[ \frac{x - \frac{\tan 2x}{2x} \times 2}{\frac{\tan x}{x}} \Bigg] = \frac{0-2(1)}{1} \\ \\ { \hspace{11.0cm} = -2}$

$\\$

$\displaystyle \text{ Question 34: } \lim \limits_{x \to 0 } \ \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{\sqrt{2} - \sqrt{1+\cos x}}{\sin^2 x} \times \frac{\sqrt{2} + \sqrt{1+\cos x}}{ \sqrt{2} + \sqrt{1+\cos x} } \Bigg]$

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

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

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

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

$\displaystyle = 2 \times \frac{1}{4} \times \frac{1}{\sqrt{2}+ \sqrt{1+\cos 0}}$

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

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$\displaystyle \text{ Question 35: } \lim \limits_{x \to 0 } \ \frac{x \tan x}{1 - \cos x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{x \tan x}{1 - \cos x} = \lim \limits_{x \to 0 } \ \frac{\frac{x\tan x}{x^2}}{\frac{1- \cos x}{x^2}} = \lim \limits_{x \to 0 } \ \frac{\frac{\tan x }{x}}{\frac{2 \sin^2 \frac{x}{2}}{\frac{x}{2} \times \frac{x}{2} \times 4} } =\frac{4}{2} = 2$

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$\displaystyle \text{ Question 36: } \lim \limits_{x \to 0 } \ \frac{x^2 + 1 - \cos x}{x \sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{x^2 + 1 - \cos x}{x \sin x} = \lim \limits_{x \to 0 } \ \frac{x^2 + 2 \sin^2 \frac{x}{2} }{x \sin x} = \lim \limits_{x \to 0 } \ \frac{1 + \frac{2 \sin^2 \frac{x}{2}}{(\frac{x}{2})^2 \times 4} }{\frac{ \sin x}{ x} } = \frac{1+\frac{1}{2}}{1} = \frac{3}{2}$

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$\displaystyle \text{ Question 37: } \lim \limits_{x \to 0 } \ \frac{\sin 2x ( \cos3x-\cos x)}{x^3}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin 2x ( \cos3x-\cos x)}{x^3} = \lim \limits_{x \to 0 } \ \frac{\sin 2x ( -2 \sin \frac{3x+x}{2} \sin \frac{3x-x}{2} )}{x^3} \\ \\ { \hspace{5.0cm} = -2 \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 2x}{2x} \times \frac{\sin 2x}{2x} \times \frac{\sin x}{x} \times 4 \Bigg] = -8}$

$\\$

$\displaystyle \text{ Question 38: } \lim \limits_{x \to 0 } \ \frac{2\sin x^{\circ} - \sin 2x^{\circ}}{x^3}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{2\sin x^{\circ} - \sin 2x^{\circ}}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2\sin (\frac{\pi x}{180}) - \sin (\frac{2\pi x}{180})}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2\sin (\frac{\pi x}{180}) - 2 \sin (\frac{\pi x}{180}) \cos(\frac{\pi x}{180}) }{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2\sin (\frac{\pi x}{180}) [1 - \cos(\frac{\pi x}{180}) ] }{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2\sin (\frac{\pi x}{180}) 2 \sin^2 (\frac{\pi x}{360} ) }{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{4\sin (\frac{\pi x}{180}) \sin^2 (\frac{\pi x}{360} ) }{\frac{\pi x}{180} \times \frac{\pi x}{180} \times \frac{\pi x}{180}} \times \frac{\pi x}{180} \times (\frac{\pi }{360})^2 \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin (\frac{\pi x}{180})}{\frac{\pi x}{180}} \times \frac{\sin(\frac{\pi x }{360}) \times \sin (\frac{\pi x }{360})}{\frac{\pi x }{360} \times \frac{\pi x }{360}} \times \frac{\pi^3}{180\times 360^2} \Bigg]$

$\displaystyle = 4 \times 1 \times 1 \times \times \frac{\pi^3}{180 \times 360 \times 360}$

$\displaystyle = (\frac{\pi }{180})^3$

$\\$

$\displaystyle \text{ Question 39: } \lim \limits_{x \to 0 } \ \frac{x^3 \cot x}{1 - \cos x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{x^3 \cot x}{1 - \cos x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{x^3 }{\tan x(1 - \cos x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{x^3 }{\tan x (2 \sin^2 \frac{x}{2} )}\Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{x}{\tan x} \times \frac{x^2}{2 \sin^2 \frac{x}{2} } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{x}{\tan x} \times \frac{\frac{x^2}{4} \times 4}{2 \sin^2 \frac{x}{2} } \Bigg]$

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

$\displaystyle = 1 \times 1 \times 2 = 2$

$\\$

$\displaystyle \text{ Question 40: } \lim \limits_{x \to 0 } \ \frac{x \tan x}{1 - \cos 2x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{x \tan x}{1 - \cos 2x} = \lim \limits_{x \to 0 } \ \frac{x \tan x}{2 \sin^2 x} = \lim \limits_{x \to 0 } \ \frac{\frac{x\tan x}{x^2}}{2 \frac{\sin^2 x^2}{x^2} } = \lim \limits_{x \to 0 } \ \frac{\frac{\tan x}{x}}{2 (\frac{\sin x}{x})^2} = \frac{1}{2}$

$\\$

$\displaystyle \text{ Question 41: } \lim \limits_{x \to 0 } \ \ \frac{\sin ( 3 + x) - \sin ( 3 - x) }{x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \frac{\sin ( 3 + x) - \sin ( 3 - x) }{x} = \lim \limits_{x \to 0 } \frac{2 \cos (\frac{3+x+3-x}{2}) \sin (\frac{3+x-3+x}{2} ) }{x} = \lim \limits_{x \to 0 } \frac{2 \cos 3 \sin x}{x} \\ \\ { \hspace{11.2cm} = 2 \cos 3 }$

$\\$

$\displaystyle \text{ Question 42: } \lim \limits_{x \to 0 } \ \frac{\cos 2x - 1}{\cos x- 1}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\cos 2x - 1}{\cos x- 1} \Bigg] \\ \\ = \lim \limits_{x \to 0 } \ \Bigg[ \frac{1 - 2 \sin^2 x-1}{1 - 2 \sin^2 \frac{x}{2} -1} \Bigg] \\ \\ = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin^2 x}{\sin^2 \frac{x}{2}} \Bigg] \\ \\ = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin^2 x}{x^2} \times \frac{x^2}{\frac{\sin^2 \frac{x}{2}}{\frac{x^2}{4} \times \frac{x^2}{4}}} \Bigg] = 4$

$\\$

$\displaystyle \text{ Question 43: } \lim \limits_{x \to 0 } \ \frac{3 \sin^2 x - 2 \sin x^2}{3x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{3 \sin^2 x - 2 \sin x^2}{3x^2} \Bigg] = \lim \limits_{x \to 0 } \ \Bigg[ \frac{3\sin^2 x}{3x^2}-\frac{2 \sin x^2}{3x^2} \Bigg] = 1 - \frac{2}{3} = \frac{1}{3}$

$\\$

$\displaystyle \text{ Question 44: } \lim \limits_{x \to 0 } \ \frac{\sqrt{1+\sin x} - \sqrt{1- \sin x}}{x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sqrt{1+\sin x} - \sqrt{1- \sin x}}{x} \Bigg]$

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

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

$\displaystyle \lim \limits_{x \to 0 } \ = \Bigg[ \frac{2\sin x}{x(\sqrt{1+\sin x} + \sqrt{1- \sin x}\ )} \Bigg]$

$\displaystyle \lim \limits_{x \to 0 } \ = \frac{2}{\sqrt{1+0}+\sqrt{1+0}} = \frac{2}{2} = 1$

$\\$

$\displaystyle \text{ Question 45: } \lim \limits_{x \to 0 } \ \frac{1 - \cos 4x}{x^2}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{1 - \cos 4x}{x^2} = 2 \lim \limits_{x \to 0 } \ \frac{2 \sin^2 2x}{x^2} = 2 \lim \limits_{x \to 0 } \ \frac{\sin 2x}{x} \times \frac{\sin 2x}{x} \\ \\ \\ { \hspace{5.0cm} = 2 \lim \limits_{x \to 0 } \ \frac{\sin 2x}{2x} \times \frac{\sin 2x}{2x} \times 4 = 2 \times 1 \times 1 \times 4 = 8 }$

$\\$

$\displaystyle \text{ Question 46: } \lim \limits_{x \to 0 } \ \frac{x \cos x + \sin x}{x^2 + \tan x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{x \cos x + \sin x}{x^2 + \tan x} \Bigg] = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\cos x + \frac{\sin x}{x}}{x + \frac{\tan x}{x}} \Bigg] = \frac{\cos 0+1}{0+1} = 2$

$\\$

$\displaystyle \text{ Question 47: } \lim \limits_{x \to 0 } \ \frac{1 - \cos 2x}{3 \tan^2 x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{1 - \cos 2x}{3 \tan^2 x} \Bigg] = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 x}{3 \tan^2 x} \Bigg] { \hspace{0.0cm} = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2}{3} \times \frac{\sin^2 x}{\sin^2 x} \times \cos^2 x \Bigg] = \frac{2}{3} \cos^2 0 = \frac{2}{3} }$

$\\$

$\displaystyle \text{ Question 48: } \lim \limits_{\theta \to 0 } \ \frac{1- \cos 4\theta}{1 - \cos 6\theta}$

Answer:

$\displaystyle \lim \limits_{\theta \to 0 } \ \frac{1- \cos 4\theta}{1 - \cos 6\theta} \\ \\ \\ = \lim \limits_{\theta \to 0 } \ \Bigg[ \frac{2 \sin^2 2\theta }{2 \sin^2 3\theta} \Bigg] \\ \\ \\ = \lim \limits_{\theta \to 0 } \ \Bigg[ \frac{\sin^2 2\theta}{(2\theta)^2} \times \frac{(2\theta)^2}{\frac{\sin^2 3\theta}{(3\theta)^2} \times (3\theta)^2 } \Bigg] \\ \\ \\ = \lim \limits_{\theta \to 0 } \ \Bigg[ \Bigg( \frac{\sin 2 \theta}{2 \theta} \Bigg)^2 \times \Bigg( \frac{3 \theta}{\sin 3 \theta} \Bigg)^2 \times \frac{4}{9} \Bigg] \\ \\ \\ = \frac{4}{9}$

$\\$

$\displaystyle \text{ Question 49: } \lim \limits_{x \to 0 } \ \frac{ax + x \cos x}{b \sin x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \frac{ax + x \cos x}{b \sin x} = \lim \limits_{x \to 0 } \ \frac{a+\cos x}{b ( \frac{\sin x}{x} )} = \frac{a+\cos 0}{b} = \frac{a+1}{b}$

$\\$

$\displaystyle \text{ Question 50: } \lim \limits_{\theta \to 0 } \ \frac{\sin 4 \theta}{\tan 3 \theta}$

Answer:

$\displaystyle \lim \limits_{\theta \to 0 } \ \frac{\sin 4 \theta}{\tan 3 \theta} = \Bigg[ \frac{\sin 4\theta}{4 \theta} \times \frac{4 \theta}{\frac{\tan 3\theta}{3\theta} \times 3 \theta } \Bigg] = \frac{4}{3}$

$\\$

$\displaystyle \text{ Question 51: } \lim \limits_{x \to 0 } \ \frac{2 \sin x - \sin 2x}{x^3}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin x - \sin 2x}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin x - 2 \sin x \cos x}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin x \times 2 \sin^2 \frac{x}{2}}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin x \times \sin^2 \frac{x}{2}}{x \times \frac{x^2}{4} } \Bigg]$

$\displaystyle = 1$

$\\$

$\displaystyle \text{ Question 52: } \lim \limits_{x \to 0 } \ \frac{1- \cos 5x}{1 - \cos 6x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{1- \cos 5x}{1 - \cos 6x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 \frac{5x}{2} }{2 \sin^2 3x } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin^2 \frac{5x}{2}}{(\frac{5x}{2})^2} \times \frac{(\frac{5x}{2})^2}{\frac{\sin^2 3x}{(3x)^2} \times (3x)^2} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \Bigg( \frac{\sin \frac{5x}{2}}{\frac{5x}{2}} \Bigg)^2 \times \frac{\frac{25}{4} x^2}{\Big( \frac{\sin 3x}{3x} \Big)^2 \times 9x^2} \Bigg]$

$\displaystyle = \frac{25}{9 \times 4} = \frac{25}{36}$

$\\$

$\displaystyle \text{ Question 53: } \lim \limits_{x \to 0 } \ \frac{\mathrm{cosec} x - \cot x}{x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\mathrm{cosec} x - \cot x}{x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{1}{\sin x}- \frac{\sin x}{\cos x} }{x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{1 - \cos x}{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 \frac{x}{2} }{x \sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 \frac{x}{2}}{(\frac{x}{2})^2} \times \frac{\frac{x^2}{4}}{\frac{x \sin x}{x^2} \times x^2 } \Bigg]$

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

$\\$

$\displaystyle \text{ Question 54: } \lim \limits_{x \to 0 } \ \frac{\sin 3x + 7x}{4x + \sin 2x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 3x + 7x}{4x + \sin 2x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin 3x}{3x} \times 3x + 7x }{4x + \frac{\sin 2x}{2x} \times 2x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin 3x}{3x} \times 3 + 7 }{4 + \frac{\sin 2x}{2x} \times 2} \Bigg]$

$\displaystyle = \frac{3+7}{4+2} = \frac{10}{6} = \frac{5}{3}$

$\\$

$\displaystyle \text{ Question 55: } \lim \limits_{x \to 0 } \ \frac{5x+4 \sin 3x}{4 \sin 2x + 7x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{5x+4 \sin 3x}{4 \sin 2x + 7x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{5x+4 \frac{\sin 3x}{3x} \times 3x}{4 \frac{\sin 2x}{2x} \times 2x + 7x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{5+4 \frac{\sin 3x}{3x} \times 3}{4 \frac{\sin 2x}{2x} \times 2 + 7} \Bigg]$

$\displaystyle = \frac{5+4 \times 3}{4 \times 2+7} = \frac{17}{15}$

$\\$

$\displaystyle \text{ Question 56: } \lim \limits_{x \to 0 } \ \frac{3 \sin x - \sin 3x}{x^3}$

Answer:

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

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{3 \sin x - (3 \sin x - 4 \sin^3 x) }{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{4 \sin^3 x }{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ 4 \Big( \frac{ \sin x }{x} \Big)^3 \Bigg]$

$\displaystyle = 4 \times 1 = 4$

$\\$

$\displaystyle \text{ Question 57: } \lim \limits_{x \to 0 } \ \frac{\tan 2x - \sin 2x}{x^3}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\tan 2x - \sin 2x}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\frac{\sin 2x}{\cos 2x} - \sin 2x}{x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 2x ( 1 - \cos 2x)}{\cos 2x \times x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 2x \cdot 2 \sin^2 x}{\cos 2x \times x^3} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin 2x}{2x} \times \frac{2}{\cos 2x} \times 2 \Bigg( \frac{\sin x}{x} \Bigg)^2 \Bigg]$

$\displaystyle = \frac{2 \times 2}{\cos 0} = 4$

$\\$

$\displaystyle \text{ Question 58: } \lim \limits_{x \to 0 } \ \frac{\sin ax + bx}{ax + \sin bx}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\sin ax + bx}{ax + \sin bx} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin ax}{ax} \times ax + bx}{ax + \frac{\sin bx}{bx} \times bx} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin ax}{ax} \times a + b}{a + \frac{\sin bx}{bx} \times b} \Bigg]$

$\displaystyle = \frac{1 \times a + b}{a+b} = 1$

$\\$

$\displaystyle \text{ Question 59: } \lim \limits_{x \to 0 } \ (\mathrm{cosec} x - \cot x)$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ [ (\mathrm{cosec} x - \cot x) ]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{1}{\sin x} - \frac{\cos x}{\sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{1- \cos x}{\sin x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{2 \sin^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \tan \frac{x}{2} \Bigg]$

$\displaystyle = 0$

$\\$

$\displaystyle \text{ Question 60: } \lim \limits_{x \to 0 } \ \frac{\Big\{ \sin (\alpha+\beta) x + \sin (\alpha - \beta)x + \sin 2x \Big\}}{\cos^2 \beta x - \cos^2 \alpha x}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\Big\{ \sin (\alpha+\beta) x + \sin (\alpha - \beta)x + \sin 2x \Big\}}{\cos^2 \beta x - \cos^2 \alpha x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{\Big\{ 2 \sin \Big( \frac{(\alpha + \beta)x +(\alpha - \beta)x}{2} \Big) \cos \Big( \frac{(\alpha + \beta)x -(\alpha - \beta)x}{2} \Big) + 2 \sin \alpha x \cos \alpha x \Big\}}{(1 - \sin^2 \beta x)-(1 - \sin^2 \alpha x)} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \sin \alpha x ( \cos \beta x + \cos \alpha x)}{\sin^2 \alpha x - \sin^2 \beta x} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \frac{\sin \alpha x}{\alpha x} \times \alpha x \times ( \cos \beta x + \cos \alpha x)}{ \frac{\sin^2 \alpha x}{\alpha^2 x^2} \times \alpha^2 x^2 - \frac{\sin^2 \beta x}{\beta^2 x^2} \times \beta^2 x^2 } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ 2 \times 1 \times \alpha x \times ( 1 + 1)}{ 1 \times \alpha^2 x^2 - 1 \times \beta^2 x^2 } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \frac{4\alpha}{\alpha^2 - \beta^2 } \times \lim \limits_{x \to 0 } \ \frac{1}{x}$

$\displaystyle = \infty$

$\\$

$\displaystyle \text{ Question 61: } \lim \limits_{x \to 0 } \ \frac{\cos ax - \cos bx}{\cos cx - 1}$

Answer:

$\displaystyle \lim \limits_{x \to 0 } \ \Bigg[ \frac{\cos ax - \cos bx}{\cos cx - 1} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{-2 \sin (\frac{ax+bx}{2}) \sin (\frac{ax-bx}{2}) }{-2 \sin^2 \frac{cx}{2}} \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin (\frac{ax+bx}{2})}{(\frac{ax+bx}{2})} \times (\frac{ax+bx}{2}) \times \frac{\sin (\frac{ax-bx}{2})}{(\frac{ax-bx}{2})} \times (\frac{ax-bx}{2}) }{ \frac{\sin^2 \frac{cx}{2}}{\frac{c^2x^2}{4}} \times \frac{c^2x^2}{4} } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{ \frac{\sin (\frac{ax+bx}{2})}{(\frac{ax+bx}{2})} \times (\frac{a+b}{2}) \times \frac{\sin (\frac{ax-bx}{2})}{(\frac{ax-bx}{2})} \times (\frac{a-b}{2}) }{ \frac{\sin^2 \frac{cx}{2}}{\frac{c^2x^2}{4}} \times \frac{c^2}{4} } \Bigg]$

$\displaystyle = \lim \limits_{x \to 0 } \ \frac{a^2-b^2}{c^2}$

$\\$

$\displaystyle \text{ Question 62: } \lim \limits_{h \to 0 } \ \frac{(a+h)^2 \sin ( a+h) - a^2 \sin a}{h}$

Answer:

$\displaystyle \lim \limits_{h \to 0 } \ \Bigg[ \frac{(a+h)^2 \sin ( a+h) - a^2 \sin a}{h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{(a^2+2ah+h^2) \sin ( a+h) - a^2 \sin a}{h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ \frac{(2ah+h^2) \sin ( a+h) + a^2 \sin ( a + h)- a^2 \sin a}{h} \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \ \Bigg[ (2a+h) \sin ( a+h) + a^2 \Bigg( \frac{ \sin ( a + h)- \sin a}{h} \Bigg) \Bigg]$

$\displaystyle = 2a \sin a + a^2 \lim \limits_{h \to 0 } \ \Bigg[ \Bigg( \frac{ 2 \cos (a+\frac{h}{2}) \sin(\frac{h}{2} ) }{h} \Bigg) \Bigg]$

$\displaystyle = 2a \sin a + a^2 \lim \limits_{h \to 0 } \ \Bigg[ \Bigg( \frac{ 2 \cos (a+\frac{h}{2}) \sin(\frac{h}{2} ) }{\frac{h}{2} \times 2} \Bigg) \Bigg]$

$\displaystyle = 2a \sin a + a^2 \cos a$

$\\$

$\displaystyle \text{ Question 63: } \lim \limits_{x \to 0 } \ kx \ \mathrm{cosec} x = \lim \limits_{x \to 0 } \ x \ \mathrm{cosec} kx, \text{ find } k.$