\displaystyle \textbf{Note:}

\displaystyle \lim \limits_{x\to0}\frac{\sin x}{x}=1 \hspace{2cm} \lim \limits_{x\to0}\cos x=1 \hspace{2cm} \lim \limits_{x\to0}\frac{\tan x}{x}=1

Evaluate the following limits:

\displaystyle \textbf{Question 1: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\sin3x}{5x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x \to 0}\frac{\sin3x}{5x}=\frac{1}{5}\lim \limits_{x \to 0}\frac{\sin3x}{3x}\times3
\displaystyle =\frac{1}{5}\times1\times3=\frac{3}{5}
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\sin x^\circ}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }x^\circ=\frac{\pi x}{180}\text{ radians,}
\displaystyle \lim \limits_{x \to 0}\frac{\sin x^\circ}{x}=\lim \limits_{x \to 0}\frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}\times\frac{\pi}{180}
\displaystyle =1\times\frac{\pi}{180}=\frac{\pi}{180}.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Evaluate } \lim \limits_{x \to 0}\frac{x^2}{\sin x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }\theta=x^2.
\displaystyle \text{When }x\to0,\text{ then }\theta\to0.
\displaystyle \therefore \lim \limits_{x \to 0}\frac{x^2}{\sin x^2}=\lim \limits_{\theta \to 0}\frac{\theta}{\sin\theta}=\lim \limits_{\theta \to 0}\frac{1}{\frac{\sin\theta}{\theta}}=\frac{1}{1}=1.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\sin x\cos x}{3x}.
\displaystyle \textbf{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\lim \limits_{x \to 0}\cos x
\displaystyle =\frac{1}{3}\times1\times1=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Evaluate } \lim \limits_{x \to 0}\frac{3\sin x-4\sin^3x}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using the identity } \sin3x=3\sin x-4\sin^3x,
\displaystyle \lim \limits_{x \to 0}\frac{3\sin x-4\sin^3x}{x}=\lim \limits_{x \to 0}\frac{\sin3x}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{\sin3x}{3x}\times3=1\times3=3.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\tan8x}{\sin2x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x \to 0}\frac{\tan8x}{\sin2x}=\lim \limits_{x \to 0}\frac{\tan8x}{8x}\times\frac{8x}{2x}\times\frac{2x}{\sin2x}
\displaystyle =1\times4\times1=4.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\tan mx}{\tan nx}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x \to 0}\frac{\tan mx}{\tan nx}=\lim \limits_{x \to 0}\frac{\tan mx}{mx}\times\frac{mx}{nx}\times\frac{nx}{\tan nx}
\displaystyle =1\times\frac{m}{n}\times1=\frac{m}{n}.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\sin5x}{\tan3x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x \to 0}\frac{\sin5x}{\tan3x}=\lim \limits_{x \to 0}\frac{\sin5x}{5x}\times\frac{5x}{3x}\times\frac{3x}{\tan3x}
\displaystyle =1\times\frac{5}{3}\times1=\frac{5}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{Evaluate } \lim \limits_{x \to 0}\frac{\sin x^\circ}{x^\circ}.
\displaystyle \textbf{Answer:}
\displaystyle \text{On direct substitution, the expression assumes the form }\frac{0}{0}.
\displaystyle \text{Since }x^\circ=\frac{\pi x}{180}\text{ radians, let }y=\frac{\pi x}{180}.
\displaystyle \text{As }x\to0,\text{ we have }y\to0.
\displaystyle \therefore \lim \limits_{x \to 0}\frac{\sin x^\circ}{x^\circ}=\lim \limits_{y \to 0}\frac{\sin y}{y}=1.
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{Evaluate } \lim \limits_{x \to 0}\frac{7x\cos x-3\sin x}{4x+\tan x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{On direct substitution, the expression assumes the form }\frac{0}{0}.
\displaystyle \text{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}}
\displaystyle =\frac{7(1)-3(1)}{4+1}=\frac{4}{5}.
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Evaluate }\lim \limits_{x\to0}\frac{\cos ax-\cos bx}{\cos cx-\cos dx}.
\displaystyle \textbf{Answer:}
\displaystyle \text{On direct substitution, the expression assumes the form }\frac{0}{0}.
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\cos ax-\cos bx}{\cos cx-\cos dx}
\displaystyle =\lim \limits_{x\to0}\frac{\sin\left(\frac{(a+b)x}{2}\right)\sin\left(\frac{(a-b)x}{2}\right)}{\sin\left(\frac{(c+d)x}{2}\right)\sin\left(\frac{(c-d)x}{2}\right)}
\displaystyle =\lim \limits_{x\to0}\Bigg[\frac{\sin\left(\frac{(a+b)x}{2}\right)}{\frac{(a+b)x}{2}}\times\frac{\sin\left(\frac{(a-b)x}{2}\right)}{\frac{(a-b)x}{2}}
\displaystyle \times\frac{\frac{(c+d)x}{2}}{\sin\left(\frac{(c+d)x}{2}\right)}\times\frac{\frac{(c-d)x}{2}}{\sin\left(\frac{(c-d)x}{2}\right)}
\displaystyle \times\frac{(a+b)(a-b)}{(c+d)(c-d)}\Bigg]
\displaystyle =1\times1\times1\times1\times\frac{(a+b)(a-b)}{(c+d)(c-d)}
\displaystyle =\frac{a^2-b^2}{c^2-d^2},\qquad c^2\ne d^2.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Evaluate }\lim \limits_{x\to0}\frac{\tan^2 3x}{x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\tan^2 3x}{x^2}=\lim \limits_{x\to0}\left(\frac{\tan3x}{3x}\right)^2\times9
\displaystyle =1^2\times9=9.
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos mx}{x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos\theta=2\sin^2\frac{\theta}{2},
\displaystyle \lim \limits_{x\to0}\frac{1-\cos mx}{x^2}=\lim \limits_{x\to0}\frac{2\sin^2\frac{mx}{2}}{x^2}
\displaystyle =2\lim \limits_{x\to0}\left(\frac{\sin\frac{mx}{2}}{\frac{mx}{2}}\right)^2\times\frac{m^2}{4}
\displaystyle =2\times1^2\times\frac{m^2}{4}=\frac{m^2}{2}.
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{Evaluate }\lim \limits_{x\to0}\frac{3\sin2x+2x}{3x+2\tan3x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{3\sin2x+2x}{3x+2\tan3x}=\lim \limits_{x\to0}\frac{3\frac{\sin2x}{x}+2}{3+2\frac{\tan3x}{x}}
\displaystyle =\lim \limits_{x\to0}\frac{6\frac{\sin2x}{2x}+2}{3+6\frac{\tan3x}{3x}}
\displaystyle =\frac{6(1)+2}{3+6(1)}=\frac{8}{9}.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Evaluate }\lim \limits_{x\to0}\frac{\cos3x-\cos7x}{x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\cos3x-\cos7x}{x^2}=\lim \limits_{x\to0}\frac{-2\sin5x\sin(-2x)}{x^2}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin5x\sin2x}{x^2}
\displaystyle =2\lim \limits_{x\to0}\frac{\sin5x}{5x}\times\frac{\sin2x}{2x}\times5\times2
\displaystyle =2\times1\times1\times5\times2=20.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{Evaluate }\lim \limits_{\theta\to0}\frac{\sin3\theta}{\tan2\theta}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{\theta\to0}\frac{\sin3\theta}{\tan2\theta}=\lim \limits_{\theta\to0}\frac{\sin3\theta}{3\theta}\times\frac{3\theta}{2\theta}\times\frac{2\theta}{\tan2\theta}
\displaystyle =1\times\frac{3}{2}\times1=\frac{3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin x^2(1-\cos x^2)}{x^6}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos\theta=2\sin^2\frac{\theta}{2},
\displaystyle \lim \limits_{x\to0}\frac{\sin x^2(1-\cos x^2)}{x^6}=\lim \limits_{x\to0}\frac{2\sin x^2\sin^2\frac{x^2}{2}}{x^6}
\displaystyle =\frac{1}{2}\lim \limits_{x\to0}\frac{\sin x^2}{x^2}\times\left(\frac{\sin\frac{x^2}{2}}{\frac{x^2}{2}}\right)^2
\displaystyle =\frac{1}{2}\times1\times1^2=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin^2 4x^2}{x^4}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\sin^2 4x^2}{x^4}=\lim \limits_{x\to0}\left(\frac{\sin4x^2}{x^2}\right)^2
\displaystyle =\lim \limits_{x\to0}\left(\frac{\sin4x^2}{4x^2}\times4\right)^2
\displaystyle =(1\times4)^2=16.
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{Evaluate }\lim \limits_{x\to0}\frac{x\cos x+2\sin x}{x^2+\tan x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{x\cos x+2\sin x}{x^2+\tan x}=\lim \limits_{x\to0}\frac{\cos x+2\frac{\sin x}{x}}{x+\frac{\tan x}{x}}
\displaystyle =\frac{1+2(1)}{0+1}=3.
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{Evaluate }\lim \limits_{x\to0}\frac{2x-\sin x}{\tan x+x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{2x-\sin x}{\tan x+x}=\lim \limits_{x\to0}\frac{2-\frac{\sin x}{x}}{\frac{\tan x}{x}+1}
\displaystyle =\frac{2-1}{1+1}=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{Evaluate }\lim \limits_{x\to0}\frac{5x\cos x+3\sin x}{3x^2+\tan x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{5x\cos x+3\sin x}{3x^2+\tan x}=\lim \limits_{x\to0}\frac{5\cos x+3\frac{\sin x}{x}}{3x+\frac{\tan x}{x}}
\displaystyle =\frac{5(1)+3(1)}{0+1}=8.
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin3x-\sin x}{\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\sin3x-\sin x}{\sin x}=\lim \limits_{x\to0}\frac{2\cos2x\sin x}{\sin x}
\displaystyle =\lim \limits_{x\to0}2\cos2x=2.
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin5x-\sin3x}{\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\sin5x-\sin3x}{\sin x}=\lim \limits_{x\to0}\frac{2\cos4x\sin x}{\sin x}
\displaystyle =\lim \limits_{x\to0}2\cos4x=2.
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{Evaluate }\lim \limits_{x\to0}\frac{\cos3x-\cos5x}{x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\cos3x-\cos5x}{x^2}=\lim \limits_{x\to0}\frac{-2\sin4x\sin(-x)}{x^2}
\displaystyle =2\lim \limits_{x\to0}\frac{\sin4x\sin x}{x^2}
\displaystyle =2\lim \limits_{x\to0}\frac{\sin4x}{4x}\times4\times\frac{\sin x}{x}
\displaystyle =2\times1\times4\times1=8.
\displaystyle \\

\displaystyle \textbf{Question 25: } \text{Evaluate }\lim \limits_{x\to0}\frac{\tan3x-2x}{3x-\sin^2x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{\tan3x-2x}{3x-\sin^2x}=\lim \limits_{x\to0}\frac{\frac{\tan3x}{x}-2}{3-\frac{\sin^2x}{x}}
\displaystyle =\lim \limits_{x\to0}\frac{3\frac{\tan3x}{3x}-2}{3-\frac{\sin x}{x}\sin x}
\displaystyle =\frac{3(1)-2}{3-(1)(0)}=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 26: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin(2+x)-\sin(2-x)}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\sin(2+x)-\sin(2-x)}{x}=\lim \limits_{x\to0}\frac{2\cos2\sin x}{x}
\displaystyle =2\cos2\lim \limits_{x\to0}\frac{\sin x}{x}=2\cos2.
\displaystyle \\

\displaystyle \textbf{Question 27: } \text{Evaluate }\lim \limits_{h\to0}\frac{(a+h)^2\sin(a+h)-a^2\sin a}{h}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{h\to0}\frac{(a+h)^2\sin(a+h)-a^2\sin a}{h}
\displaystyle =\lim \limits_{h\to0}\frac{(a^2+2ah+h^2)\sin(a+h)-a^2\sin a}{h}
\displaystyle =\lim \limits_{h\to0}\left[a^2\frac{\sin(a+h)-\sin a}{h}+2a\sin(a+h)+h\sin(a+h)\right]
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \frac{\sin(a+h)-\sin a}{h}=\cos\left(a+\frac{h}{2}\right)\frac{\sin\frac{h}{2}}{\frac{h}{2}}
\displaystyle \therefore \lim \limits_{h\to0}\frac{(a+h)^2\sin(a+h)-a^2\sin a}{h}
\displaystyle =a^2\cos a\cdot1+2a\sin a+0
\displaystyle =a^2\cos a+2a\sin a.
\displaystyle \\

\displaystyle \textbf{Question 28: } \text{Evaluate }\lim \limits_{x\to0}\frac{\tan x-\sin x}{\sin3x-3\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\tan x-\sin x}{\sin3x-3\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{\sin x}{\cos x}-\sin x}{3\sin x-4\sin^3x-3\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{\sin x(1-\cos x)}{-4\cos x\sin^3x}
\displaystyle =-\frac{1}{4}\lim \limits_{x\to0}\frac{1-\cos x}{\cos x\sin^2x}
\displaystyle =-\frac{1}{4}\lim \limits_{x\to0}\frac{2\sin^2\frac{x}{2}}{\cos x\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)^2}
\displaystyle =-\frac{1}{8}\lim \limits_{x\to0}\frac{1}{\cos x\cos^2\frac{x}{2}}
\displaystyle =-\frac{1}{8}.
\displaystyle \\

\displaystyle \textbf{Question 29: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sec5x-\sec3x}{\sec3x-\sec x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\sec5x-\sec3x}{\sec3x-\sec x}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{1}{\cos5x}-\frac{1}{\cos3x}}{\frac{1}{\cos3x}-\frac{1}{\cos x}}
\displaystyle =\lim \limits_{x\to0}\frac{(\cos3x-\cos5x)\cos x}{\cos5x(\cos x-\cos3x)}
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle =\lim \limits_{x\to0}\frac{2\sin4x\sin x\cos x}{2\cos5x\sin2x\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{\sin4x}{\sin2x}\times\frac{\cos x}{\cos5x}
\displaystyle =\lim \limits_{x\to0}\frac{\sin4x}{4x}\times\frac{4x}{2x}\times\frac{2x}{\sin2x}\times\frac{\cos x}{\cos5x}
\displaystyle =1\times2\times1\times1=2.
\displaystyle \\

\displaystyle \textbf{Question 30: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos2x}{\cos2x-\cos8x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos2x=2\sin^2x,
\displaystyle \text{and }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{1-\cos2x}{\cos2x-\cos8x}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin^2x}{-2\sin5x\sin(-3x)}
\displaystyle =\lim \limits_{x\to0}\frac{\sin^2x}{\sin5x\sin3x}
\displaystyle =\lim \limits_{x\to0}\frac{\sin x}{x}\times\frac{\sin x}{x}\times\frac{5x}{\sin5x}\times\frac{3x}{\sin3x}\times\frac{1}{15}
\displaystyle =1\times1\times1\times1\times\frac{1}{15}=\frac{1}{15}.
\displaystyle \\

\displaystyle \textbf{Question 31: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos2x+\tan^2x}{x\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos2x=2\sin^2x,
\displaystyle \lim \limits_{x\to0}\frac{1-\cos2x+\tan^2x}{x\sin x}=\lim \limits_{x\to0}\frac{2\sin^2x+\tan^2x}{x\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{2\left(\frac{\sin x}{x}\right)^2+\left(\frac{\tan x}{x}\right)^2}{\frac{\sin x}{x}}
\displaystyle =\frac{2(1)^2+(1)^2}{1}=3.
\displaystyle \\

\displaystyle \textbf{Question 32: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin(a+x)+\sin(a-x)-2\sin a}{x\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
\displaystyle \sin(a+x)+\sin(a-x)=2\sin a\cos x.
\displaystyle \therefore \lim \limits_{x\to0}\frac{\sin(a+x)+\sin(a-x)-2\sin a}{x\sin x}
\displaystyle =2\sin a\lim \limits_{x\to0}\frac{\cos x-1}{x\sin x}
\displaystyle =-4\sin a\lim \limits_{x\to0}\frac{\sin^2\frac{x}{2}}{x\sin x}
\displaystyle =-\sin a\lim \limits_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\frac{x}{\sin x}
\displaystyle =-\sin a\times1^2\times1=-\sin a.
\displaystyle \\

\displaystyle \textbf{Question 33: } \text{Evaluate }\lim \limits_{x\to0}\frac{x^2-\tan2x}{\tan x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{x^2-\tan2x}{\tan x}=\lim \limits_{x\to0}\frac{x-\frac{\tan2x}{x}}{\frac{\tan x}{x}}
\displaystyle =\lim \limits_{x\to0}\frac{x-2\frac{\tan2x}{2x}}{\frac{\tan x}{x}}
\displaystyle =\frac{0-2(1)}{1}=-2.
\displaystyle \\

\displaystyle \textbf{Question 34: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin^2x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Rationalizing the numerator,}
\displaystyle \lim \limits_{x\to0}\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin^2x}
\displaystyle =\lim \limits_{x\to0}\frac{\left(\sqrt{2}-\sqrt{1+\cos x}\right)\left(\sqrt{2}+\sqrt{1+\cos x}\right)}{\sin^2x\left(\sqrt{2}+\sqrt{1+\cos x}\right)}
\displaystyle =\lim \limits_{x\to0}\frac{1-\cos x}{\sin^2x\left(\sqrt{2}+\sqrt{1+\cos x}\right)}
\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2}\text{ and }\sin x=2\sin\frac{x}{2}\cos\frac{x}{2},
\displaystyle =\lim \limits_{x\to0}\frac{2\sin^2\frac{x}{2}}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}\left(\sqrt{2}+\sqrt{1+\cos x}\right)}
\displaystyle =\lim \limits_{x\to0}\frac{1}{2\cos^2\frac{x}{2}\left(\sqrt{2}+\sqrt{1+\cos x}\right)}
\displaystyle =\frac{1}{2(1)\left(\sqrt{2}+\sqrt{2}\right)}=\frac{1}{4\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 35: } \text{Evaluate }\lim \limits_{x\to0}\frac{x\tan x}{1-\cos x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{x\tan x}{1-\cos x}=\lim \limits_{x\to0}\frac{\frac{\tan x}{x}}{\frac{1-\cos x}{x^2}}
\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2},
\displaystyle =\lim \limits_{x\to0}\frac{\frac{\tan x}{x}}{\frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2}
\displaystyle =\frac{1}{\frac{1}{2}(1)^2}=2.
\displaystyle \\

\displaystyle \textbf{Question 36: } \text{Evaluate }\lim \limits_{x\to0}\frac{x^2+1-\cos x}{x\sin x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2},
\displaystyle \lim \limits_{x\to0}\frac{x^2+1-\cos x}{x\sin x}=\lim \limits_{x\to0}\frac{x^2+2\sin^2\frac{x}{2}}{x\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{1+\frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2}{\frac{\sin x}{x}}
\displaystyle =\frac{1+\frac{1}{2}(1)^2}{1}=\frac{3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 37: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin2x(\cos3x-\cos x)}{x^3}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \cos3x-\cos x=-2\sin2x\sin x.
\displaystyle \therefore \lim \limits_{x\to0}\frac{\sin2x(\cos3x-\cos x)}{x^3}=-2\lim \limits_{x\to0}\frac{\sin^22x\sin x}{x^3}
\displaystyle =-8\lim \limits_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\frac{\sin x}{x}
\displaystyle =-8(1)^2(1)=-8.
\displaystyle \\

\displaystyle \textbf{Question 38: } \text{Evaluate }\lim \limits_{x\to0}\frac{2\sin x^\circ-\sin2x^\circ}{x^3}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }x^\circ=\frac{\pi x}{180}\text{ radians, let }a=\frac{\pi}{180}.
\displaystyle \lim \limits_{x\to0}\frac{2\sin x^\circ-\sin2x^\circ}{x^3}=\lim \limits_{x\to0}\frac{2\sin ax-\sin2ax}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin ax-2\sin ax\cos ax}{x^3}
\displaystyle =2\lim \limits_{x\to0}\frac{\sin ax(1-\cos ax)}{x^3}
\displaystyle =4\lim \limits_{x\to0}\frac{\sin ax\sin^2\frac{ax}{2}}{x^3}
\displaystyle =4\lim \limits_{x\to0}\frac{\sin ax}{ax}\left(\frac{\sin\frac{ax}{2}}{\frac{ax}{2}}\right)^2\times a\times\frac{a^2}{4}
\displaystyle =4\times1\times1^2\times\frac{a^3}{4}=a^3
\displaystyle =\left(\frac{\pi}{180}\right)^3.
\displaystyle \\

\displaystyle \textbf{Question 39: } \text{Evaluate }\lim \limits_{x\to0}\frac{x^3\cot x}{1-\cos x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{x^3\cot x}{1-\cos x}=\lim \limits_{x\to0}\frac{x^3}{\tan x(1-\cos x)}
\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2},
\displaystyle =\lim \limits_{x\to0}\frac{x^3}{2\tan x\sin^2\frac{x}{2}}
\displaystyle =\lim \limits_{x\to0}\frac{x}{\tan x}\times\frac{x^2}{2\sin^2\frac{x}{2}}
\displaystyle =2\lim \limits_{x\to0}\frac{x}{\tan x}\left(\frac{\frac{x}{2}}{\sin\frac{x}{2}}\right)^2
\displaystyle =2\times1\times1^2=2.
\displaystyle \\

\displaystyle \textbf{Question 40: } \text{Evaluate }\lim \limits_{x\to0}\frac{x\tan x}{1-\cos2x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos2x=2\sin^2x,
\displaystyle \lim \limits_{x\to0}\frac{x\tan x}{1-\cos2x}=\lim \limits_{x\to0}\frac{x\tan x}{2\sin^2x}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{\tan x}{x}}{2\left(\frac{\sin x}{x}\right)^2}
\displaystyle =\frac{1}{2(1)^2}=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 41: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin(3+x)-\sin(3-x)}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \lim \limits_{x\to0}\frac{\sin(3+x)-\sin(3-x)}{x}
\displaystyle =\lim \limits_{x\to0}\frac{2\cos3\sin x}{x}
\displaystyle =2\cos3\lim \limits_{x\to0}\frac{\sin x}{x}=2\cos3.
\displaystyle \\

\displaystyle \textbf{Question 42: } \text{Evaluate }\lim \limits_{x\to0}\frac{\cos2x-1}{\cos x-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\cos2x-1=-2\sin^2x\text{ and }\cos x-1=-2\sin^2\frac{x}{2},
\displaystyle \lim \limits_{x\to0}\frac{\cos2x-1}{\cos x-1}=\lim \limits_{x\to0}\frac{\sin^2x}{\sin^2\frac{x}{2}}
\displaystyle =\lim \limits_{x\to0}\left(\frac{\sin x}{\sin\frac{x}{2}}\right)^2
\displaystyle =\lim \limits_{x\to0}\left(2\cos\frac{x}{2}\right)^2
\displaystyle =4\cos^2 0=4.
\displaystyle \\

\displaystyle \textbf{Question 43: } \text{Evaluate }\lim \limits_{x\to0}\frac{3\sin^2x-2\sin x^2}{3x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{3\sin^2x-2\sin x^2}{3x^2}
\displaystyle =\lim \limits_{x\to0}\left[\frac{\sin^2x}{x^2}-\frac{2}{3}\frac{\sin x^2}{x^2}\right]
\displaystyle =\lim \limits_{x\to0}\left[\left(\frac{\sin x}{x}\right)^2-\frac{2}{3}\frac{\sin x^2}{x^2}\right]
\displaystyle =1^2-\frac{2}{3}(1)=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 44: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Rationalizing the numerator,}
\displaystyle \lim \limits_{x\to0}\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}
\displaystyle =\lim \limits_{x\to0}\frac{\left(\sqrt{1+\sin x}-\sqrt{1-\sin x}\right)\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}{x\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin x}{x\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}
\displaystyle =\lim \limits_{x\to0}\frac{2\frac{\sin x}{x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}
\displaystyle =\frac{2(1)}{\sqrt{1}+\sqrt{1}}=1.
\displaystyle \\

\displaystyle \textbf{Question 45: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos4x}{x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos4x=2\sin^22x,
\displaystyle \lim \limits_{x\to0}\frac{1-\cos4x}{x^2}=2\lim \limits_{x\to0}\frac{\sin^22x}{x^2}
\displaystyle =2\lim \limits_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\times4
\displaystyle =2\times1^2\times4=8.
\displaystyle \\

\displaystyle \textbf{Question 46: } \text{Evaluate }\lim \limits_{x\to0}\frac{x\cos x+\sin x}{x^2+\tan x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle \lim \limits_{x\to0}\frac{x\cos x+\sin x}{x^2+\tan x}
\displaystyle =\lim \limits_{x\to0}\frac{\cos x+\frac{\sin x}{x}}{x+\frac{\tan x}{x}}
\displaystyle =\frac{\cos0+1}{0+1}=2.
\displaystyle \\

\displaystyle \textbf{Question 47: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos2x}{3\tan^2x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos2x=2\sin^2x,
\displaystyle \lim \limits_{x\to0}\frac{1-\cos2x}{3\tan^2x}=\lim \limits_{x\to0}\frac{2\sin^2x}{3\tan^2x}
\displaystyle =\frac{2}{3}\lim \limits_{x\to0}\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}
\displaystyle =\frac{2}{3}\lim \limits_{x\to0}\cos^2x=\frac{2}{3}\cos^20=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 48: } \text{Evaluate }\lim \limits_{\theta\to0}\frac{1-\cos4\theta}{1-\cos6\theta}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos2A=2\sin^2A,
\displaystyle \lim \limits_{\theta\to0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim \limits_{\theta\to0}\frac{2\sin^22\theta}{2\sin^23\theta}
\displaystyle =\lim \limits_{\theta\to0}\left(\frac{\sin2\theta}{2\theta}\right)^2\left(\frac{3\theta}{\sin3\theta}\right)^2\frac{4}{9}
\displaystyle =1^2\times1^2\times\frac{4}{9}=\frac{4}{9}.
\displaystyle \\

\displaystyle \textbf{Question 49: } \text{Evaluate }\lim \limits_{x\to0}\frac{ax+x\cos x}{b\sin x},\quad b\neq0.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{ax+x\cos x}{b\sin x}=\lim \limits_{x\to0}\frac{x(a+\cos x)}{b\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{a+\cos x}{b\left(\frac{\sin x}{x}\right)}
\displaystyle =\frac{a+\cos0}{b(1)}=\frac{a+1}{b}.
\displaystyle \\

\displaystyle \textbf{Question 50: } \text{Evaluate }\lim \limits_{\theta\to0}\frac{\sin4\theta}{\tan3\theta}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{\theta\to0}\frac{\sin4\theta}{\tan3\theta}
\displaystyle =\lim \limits_{\theta\to0}\frac{\sin4\theta}{4\theta}\times\frac{3\theta}{\tan3\theta}\times\frac{4}{3}
\displaystyle =1\times1\times\frac{4}{3}=\frac{4}{3}.
\displaystyle \\

\displaystyle \textbf{Question 51: } \text{Evaluate }\lim \limits_{x\to0}\frac{2\sin x-\sin2x}{x^3}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{2\sin x-\sin2x}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin x-2\sin x\cos x}{x^3}
\displaystyle =2\lim \limits_{x\to0}\frac{\sin x(1-\cos x)}{x^3}
\displaystyle =4\lim \limits_{x\to0}\frac{\sin x\sin^2\frac{x}{2}}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{\sin x}{x}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2
\displaystyle =1\times1^2=1.
\displaystyle \\

\displaystyle \textbf{Question 52: } \text{Evaluate }\lim \limits_{x\to0}\frac{1-\cos5x}{1-\cos6x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }1-\cos A=2\sin^2\frac{A}{2},
\displaystyle \lim \limits_{x\to0}\frac{1-\cos5x}{1-\cos6x}=\lim \limits_{x\to0}\frac{2\sin^2\frac{5x}{2}}{2\sin^23x}
\displaystyle =\lim \limits_{x\to0}\left(\frac{\sin\frac{5x}{2}}{\frac{5x}{2}}\right)^2\left(\frac{3x}{\sin3x}\right)^2\frac{25}{36}
\displaystyle =1^2\times1^2\times\frac{25}{36}=\frac{25}{36}.
\displaystyle \\

\displaystyle \textbf{Question 53: } \text{Evaluate }\lim \limits_{x\to0}\frac{\mathrm{cosec}\,x-\cot x}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\mathrm{cosec}\,x-\cot x}{x}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}{x}
\displaystyle =\lim \limits_{x\to0}\frac{1-\cos x}{x\sin x}
\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2},
\displaystyle =\lim \limits_{x\to0}\frac{2\sin^2\frac{x}{2}}{x\sin x}
\displaystyle =\frac{1}{2}\lim \limits_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\frac{x}{\sin x}
\displaystyle =\frac{1}{2}\times1^2\times1=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 54: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin3x+7x}{4x+\sin2x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\sin3x+7x}{4x+\sin2x}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{\sin3x}{3x}\times3x+7x}{4x+\frac{\sin2x}{2x}\times2x}
\displaystyle =\lim \limits_{x\to0}\frac{3\frac{\sin3x}{3x}+7}{4+2\frac{\sin2x}{2x}}
\displaystyle =\frac{3(1)+7}{4+2(1)}=\frac{10}{6}=\frac{5}{3}.
\displaystyle \\

\displaystyle \textbf{Question 55: } \text{Evaluate }\lim \limits_{x\to0}\frac{5x+4\sin3x}{4\sin2x+7x}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{5x+4\sin3x}{4\sin2x+7x}
\displaystyle =\lim \limits_{x\to0}\frac{5x+4\left(\frac{\sin3x}{3x}\right)3x}{4\left(\frac{\sin2x}{2x}\right)2x+7x}
\displaystyle =\lim \limits_{x\to0}\frac{5+12\left(\frac{\sin3x}{3x}\right)}{8\left(\frac{\sin2x}{2x}\right)+7}
\displaystyle =\frac{5+12(1)}{8(1)+7}=\frac{17}{15}.
\displaystyle \\

\displaystyle \textbf{Question 56: } \text{Evaluate }\lim \limits_{x\to0}\frac{3\sin x-\sin3x}{x^3}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{3\sin x-\sin3x}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{3\sin x-\left(3\sin x-4\sin^3x\right)}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{4\sin^3x}{x^3}
\displaystyle =4\lim \limits_{x\to0}\left(\frac{\sin x}{x}\right)^3
\displaystyle =4(1)^3=4.
\displaystyle \\

\displaystyle \textbf{Question 57: } \text{Evaluate }\lim \limits_{x\to0}\frac{\tan2x-\sin2x}{x^3}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\frac{\tan2x-\sin2x}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{\frac{\sin2x}{\cos2x}-\sin2x}{x^3}
\displaystyle =\lim \limits_{x\to0}\frac{\sin2x(1-\cos2x)}{\cos2x\;x^3}
\displaystyle =\lim \limits_{x\to0}\frac{\sin2x\cdot2\sin^2x}{\cos2x\;x^3}
\displaystyle =\lim \limits_{x\to0}\left(\frac{\sin2x}{2x}\right)\left(\frac{2}{\cos2x}\right)\left(\frac{\sin x}{x}\right)^2
\displaystyle =1\times2\times1^2=2
\displaystyle \text{Since }\frac{\sin2x}{x}=2\left(\frac{\sin2x}{2x}\right),
\displaystyle \therefore \lim \limits_{x\to0}\frac{\tan2x-\sin2x}{x^3}=2\times2\times1^2=4.
\displaystyle \\

\displaystyle \textbf{Question 58: } \text{Evaluate }\lim \limits_{x\to0}\frac{\sin ax+bx}{ax+\sin bx}.
\displaystyle \textbf{Answer:}
\displaystyle \text{For }a+b\neq0,
\displaystyle \lim \limits_{x\to0}\frac{\sin ax+bx}{ax+\sin bx}
\displaystyle =\lim \limits_{x\to0}\frac{a\left(\frac{\sin ax}{ax}\right)+b}{a+b\left(\frac{\sin bx}{bx}\right)}
\displaystyle =\frac{a(1)+b}{a+b(1)}=\frac{a+b}{a+b}=1.
\displaystyle \text{If }a+b=0\text{ and }a\neq0,\text{ then }b=-a,
\displaystyle \frac{\sin ax+bx}{ax+\sin bx}=\frac{\sin ax-ax}{ax-\sin ax}=-1.
\displaystyle \therefore \lim \limits_{x\to0}\frac{\sin ax+bx}{ax+\sin bx}=\begin{cases}1,&a+b\neq0,\\-1,&a+b=0,\ a\neq0.\end{cases}
\displaystyle \\

\displaystyle \textbf{Question 59: } \text{Evaluate }\lim \limits_{x\to0}\left(\mathrm{cosec}\,x-\cot x\right).
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}\left(\mathrm{cosec}\,x-\cot x\right)
\displaystyle =\lim \limits_{x\to0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)
\displaystyle =\lim \limits_{x\to0}\frac{1-\cos x}{\sin x}
\displaystyle =\lim \limits_{x\to0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}
\displaystyle =\lim \limits_{x\to0}\tan\frac{x}{2}=0.
\displaystyle \\

\displaystyle \textbf{Question 60: } \text{Evaluate }\lim \limits_{x\to0}\frac{\cos ax-\cos bx}{\cos cx-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \text{and }\cos cx-1=-2\sin^2\frac{cx}{2},
\displaystyle \lim \limits_{x\to0}\frac{\cos ax-\cos bx}{\cos cx-1}
\displaystyle =\lim \limits_{x\to0}\frac{\sin\frac{(a+b)x}{2}\sin\frac{(a-b)x}{2}}{\sin^2\frac{cx}{2}}
\displaystyle =\lim \limits_{x\to0}\frac{\sin\frac{(a+b)x}{2}}{\frac{(a+b)x}{2}}\times\frac{\sin\frac{(a-b)x}{2}}{\frac{(a-b)x}{2}}
\displaystyle \hspace{1.5cm}\times\left(\frac{\frac{cx}{2}}{\sin\frac{cx}{2}}\right)^2\times\frac{(a+b)(a-b)}{c^2}
\displaystyle =1\times1\times1^2\times\frac{a^2-b^2}{c^2}
\displaystyle =\frac{a^2-b^2}{c^2},\quad c\neq0.
\displaystyle \\

\displaystyle \textbf{Question 61: } \text{Evaluate }\lim \limits_{h\to0}\frac{(a+h)^2\sin(a+h)-a^2\sin a}{h}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{h\to0}\frac{(a+h)^2\sin(a+h)-a^2\sin a}{h}
\displaystyle =\lim \limits_{h\to0}\frac{(a^2+2ah+h^2)\sin(a+h)-a^2\sin a}{h}
\displaystyle =\lim \limits_{h\to0}\left[(2a+h)\sin(a+h)+a^2\frac{\sin(a+h)-\sin a}{h}\right]
\displaystyle =2a\sin a+a^2\lim \limits_{h\to0}\frac{\sin(a+h)-\sin a}{h}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle =2a\sin a+a^2\lim \limits_{h\to0}\left[\cos\left(a+\frac{h}{2}\right)\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right]
\displaystyle =2a\sin a+a^2\cos a.
\displaystyle \\

\displaystyle \textbf{Question 62: } \text{If }\lim \limits_{x\to0}kx\,\mathrm{cosec}\,x=\lim \limits_{x\to0}x\,\mathrm{cosec}\,kx,\text{ find }k.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to0}kx\,\mathrm{cosec}\,x=\lim \limits_{x\to0}x\,\mathrm{cosec}\,kx
\displaystyle \Rightarrow \lim \limits_{x\to0}\frac{kx}{\sin x}=\lim \limits_{x\to0}\frac{x}{\sin kx}
\displaystyle \Rightarrow \lim \limits_{x\to0}\frac{k}{\frac{\sin x}{x}}=\lim \limits_{x\to0}\frac{1}{k\left(\frac{\sin kx}{kx}\right)}
\displaystyle \Rightarrow k=\frac{1}{k}
\displaystyle \Rightarrow k^2=1
\displaystyle \therefore k=\pm1.
\displaystyle \\

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