Note:

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

Evaluate the following limits

\displaystyle \textbf{Question 1: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)\tan x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\frac{\pi}{2}-x.
\displaystyle \text{As }x\to\frac{\pi}{2},\ y\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)\tan x
\displaystyle =\lim \limits_{y\to0}y\tan\left(\frac{\pi}{2}-y\right)
\displaystyle =\lim \limits_{y\to0}y\cot y
\displaystyle =\lim \limits_{y\to0}y\left(\frac{\cos y}{\sin y}\right)
\displaystyle =\lim \limits_{y\to0}\frac{\cos y}{\frac{\sin y}{y}}
\displaystyle =\frac{1}{1}=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{\sin2x}{\cos x}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{\sin2x}{\cos x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{2\sin x\cos x}{\cos x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}2\sin x
\displaystyle =2\times1=2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{\cos^2x}{1-\sin x}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{\cos^2x}{1-\sin x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{1-\sin^2x}{1-\sin x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{(1-\sin x)(1+\sin x)}{1-\sin x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}(1+\sin x)
\displaystyle =1+1=2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{3}}\frac{\sqrt{1-\cos6x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{3}}\frac{\sqrt{1-\cos6x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{3}}\frac{\sqrt{2\sin^23x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{3}}\frac{|\sin3x|}{\frac{\pi}{3}-x}
\displaystyle \text{Let }x=\frac{\pi}{3}+h,\text{ so that }h\to0.
\displaystyle \frac{|\sin3x|}{\frac{\pi}{3}-x}=\frac{|\sin(\pi+3h)|}{-h}
\displaystyle =\frac{|-\sin3h|}{-h}=-\frac{|\sin3h|}{h}.
\displaystyle \text{For }h\to0^{-},
\displaystyle -\frac{|\sin3h|}{h}=\frac{\sin3h}{h}
\displaystyle =3\left(\frac{\sin3h}{3h}\right)\to3.
\displaystyle \therefore \lim \limits_{x\to\left(\frac{\pi}{3}\right)^{-}}\frac{\sqrt{1-\cos6x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=3.
\displaystyle \text{For }h\to0^{+},
\displaystyle -\frac{|\sin3h|}{h}=-\frac{\sin3h}{h}
\displaystyle =-3\left(\frac{\sin3h}{3h}\right)\to-3.
\displaystyle \therefore \lim \limits_{x\to\left(\frac{\pi}{3}\right)^{+}}\frac{\sqrt{1-\cos6x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=-3.
\displaystyle \text{Since the left-hand and right-hand limits are unequal, the limit does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Evaluate }\lim \limits_{x\to a}\frac{\cos x-\cos a}{x-a}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to a}\frac{\cos x-\cos a}{x-a}
\displaystyle =\lim \limits_{x\to a}\frac{-2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)}{x-a}
\displaystyle =\lim \limits_{x\to a}\left[-\sin\left(\frac{x+a}{2}\right)\frac{\sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}\right]
\displaystyle =-\sin\left(\frac{a+a}{2}\right)\times1
\displaystyle =-\sin a.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{4}+h.
\displaystyle \text{As }x\to\frac{\pi}{4},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}
\displaystyle =\lim \limits_{h\to0}\frac{1-\tan\left(\frac{\pi}{4}+h\right)}{h}
\displaystyle =\lim \limits_{h\to0}\frac{1-\frac{1+\tan h}{1-\tan h}}{h}
\displaystyle =\lim \limits_{h\to0}\frac{1-\tan h-1-\tan h}{h(1-\tan h)}
\displaystyle =\lim \limits_{h\to0}\left[-\frac{2\tan h}{h(1-\tan h)}\right]
\displaystyle =-2\lim \limits_{h\to0}\left(\frac{\tan h}{h}\right)\frac{1}{1-\tan h}
\displaystyle =-2\times1\times\frac{1}{1-0}=-2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{1-\sin x}{\left(\frac{\pi}{2}-x\right)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=\frac{\pi}{2}-x.
\displaystyle \text{As }x\to\frac{\pi}{2},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{1-\sin x}{\left(\frac{\pi}{2}-x\right)^2}
\displaystyle =\lim \limits_{h\to0}\frac{1-\sin\left(\frac{\pi}{2}-h\right)}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos h}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2\left(\frac{h}{2}\right)}{h^2}
\displaystyle =\frac{1}{2}\lim \limits_{h\to0}\left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2
\displaystyle =\frac{1}{2}\times1^2=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{3}}\frac{\sqrt{3}-\tan x}{\pi-3x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{3}-h.
\displaystyle \text{As }x\to\frac{\pi}{3},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{3}}\frac{\sqrt{3}-\tan x}{\pi-3x}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{3}-\tan\left(\frac{\pi}{3}-h\right)}{3h}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{3}-\frac{\sqrt{3}-\tan h}{1+\sqrt{3}\tan h}}{3h}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{3}(1+\sqrt{3}\tan h)-(\sqrt{3}-\tan h)}{3h(1+\sqrt{3}\tan h)}
\displaystyle =\lim \limits_{h\to0}\frac{4\tan h}{3h(1+\sqrt{3}\tan h)}
\displaystyle =\frac{4}{3}\lim \limits_{h\to0}\left(\frac{\tan h}{h}\right)\frac{1}{1+\sqrt{3}\tan h}
\displaystyle =\frac{4}{3}\times1\times1=\frac{4}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Evaluate }\lim \limits_{x\to a}\frac{a\sin x-x\sin a}{ax^2-xa^2},\ a\neq0.
\displaystyle \text{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,\text{ so that }x=a+t\text{ and }t\to0.
\displaystyle =\lim \limits_{t\to0}\frac{a\sin(a+t)-(a+t)\sin a}{a(a+t)t}
\displaystyle =\lim \limits_{t\to0}\frac{a\sin t\cos a+a\sin a\cos t-a\sin a-t\sin a}{a(a+t)t}
\displaystyle =\lim \limits_{t\to0}\frac{a\sin t\cos a+a\sin a(\cos t-1)-t\sin a}{a(a+t)t}
\displaystyle =\lim \limits_{t\to0}\frac{a\sin t\cos a-2a\sin a\sin^2\left(\frac{t}{2}\right)-t\sin a}{a(a+t)t}
\displaystyle =\lim \limits_{t\to0}\frac{\sin t}{t}\frac{\cos a}{a+t}
\displaystyle \quad-\lim \limits_{t\to0}\frac{2\sin a\sin^2\left(\frac{t}{2}\right)}{(a+t)t}
\displaystyle \quad-\lim \limits_{t\to0}\frac{\sin a}{a(a+t)}
\displaystyle =\frac{\cos a}{a}-0-\frac{\sin a}{a^2}
\displaystyle =\frac{a\cos a-\sin a}{a^2}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos^2x}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos^2x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\left[\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos^2x}\times\frac{\sqrt{2}+\sqrt{1+\sin x}}{\sqrt{2}+\sqrt{1+\sin x}}\right]
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{2-(1+\sin x)}{\cos^2x\left(\sqrt{2}+\sqrt{1+\sin x}\right)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{1-\sin x}{(1-\sin^2x)\left(\sqrt{2}+\sqrt{1+\sin x}\right)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{1}{(1+\sin x)\left(\sqrt{2}+\sqrt{1+\sin x}\right)}
\displaystyle =\frac{1}{(1+1)(\sqrt{2}+\sqrt{2})}
\displaystyle =\frac{1}{4\sqrt{2}}=\frac{\sqrt{2}}{8}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{\sqrt{2-\sin x}-1}{\left(\frac{\pi}{2}-x\right)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=\frac{\pi}{2}-x.
\displaystyle \text{As }x\to\frac{\pi}{2},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{\sqrt{2-\sin x}-1}{\left(\frac{\pi}{2}-x\right)^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2-\sin\left(\frac{\pi}{2}-h\right)}-1}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2-\cos h}-1}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{(\sqrt{2-\cos h}-1)(\sqrt{2-\cos h}+1)}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos h}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2\left(\frac{h}{2}\right)}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\frac{1}{2}\lim \limits_{h\to0}\left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2
\displaystyle \qquad\times\lim \limits_{h\to0}\frac{1}{\sqrt{2-\cos h}+1}
\displaystyle =\frac{1}{2}\times1\times\frac{1}{\sqrt{2-1}+1}
\displaystyle =\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{\left(\frac{\pi}{4}-x\right)^2}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{\left(\frac{\pi}{4}-x\right)^2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{1-\left(\frac{\cos x}{\sqrt{2}}+\frac{\sin x}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}\left(\frac{\pi}{4}-x\right)^2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\left[1-\left(\cos\frac{\pi}{4}\cos x+\sin\frac{\pi}{4}\sin x\right)\right]}{\left(\frac{\pi}{4}-x\right)^2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\left[1-\cos\left(\frac{\pi}{4}-x\right)\right]}{\left(\frac{\pi}{4}-x\right)^2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{2\sqrt{2}\sin^2\left(\frac{\frac{\pi}{4}-x}{2}\right)}{\left(\frac{\pi}{4}-x\right)^2}
\displaystyle =\frac{\sqrt{2}}{2}\lim \limits_{x\to\frac{\pi}{4}}\left[\frac{\sin\left(\frac{\frac{\pi}{4}-x}{2}\right)}{\frac{\frac{\pi}{4}-x}{2}}\right]^2
\displaystyle =\frac{\sqrt{2}}{2}\times1=\frac{1}{\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{8}}\frac{\cot4x-\cos4x}{(\pi-8x)^3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=\frac{\pi}{8}-x.
\displaystyle \text{As }x\to\frac{\pi}{8},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{8}}\frac{\cot4x-\cos4x}{(\pi-8x)^3}
\displaystyle =\lim \limits_{h\to0}\frac{\cot4\left(\frac{\pi}{8}-h\right)-\cos4\left(\frac{\pi}{8}-h\right)}{\left[\pi-8\left(\frac{\pi}{8}-h\right)\right]^3}
\displaystyle =\lim \limits_{h\to0}\frac{\cot\left(\frac{\pi}{2}-4h\right)-\cos\left(\frac{\pi}{2}-4h\right)}{(8h)^3}
\displaystyle =\lim \limits_{h\to0}\frac{\tan4h-\sin4h}{512h^3}
\displaystyle =\lim \limits_{h\to0}\frac{\sin4h(1-\cos4h)}{512h^3\cos4h}
\displaystyle =\lim \limits_{h\to0}\frac{\tan4h\cdot2\sin^22h}{512h^3}
\displaystyle =\frac{1}{16}\lim \limits_{h\to0}\left(\frac{\tan4h}{4h}\right)\left(\frac{\sin2h}{2h}\right)^2
\displaystyle =\frac{1}{16}\times1\times1
\displaystyle =\frac{1}{16}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Evaluate }\lim \limits_{x\to a}\frac{\cos x-\cos a}{\sqrt{x}-\sqrt{a}},\quad a>0.
\displaystyle \text{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\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)}{\sqrt{x}-\sqrt{a}}
\displaystyle =\lim \limits_{x\to a}\frac{-2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)(\sqrt{x}+\sqrt{a})}{x-a}
\displaystyle =-\lim \limits_{x\to a}\sin\left(\frac{x+a}{2}\right)\frac{\sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}(\sqrt{x}+\sqrt{a})
\displaystyle =-\sin\left(\frac{a+a}{2}\right)\times1\times(\sqrt{a}+\sqrt{a})
\displaystyle =-\sin a\times2\sqrt{a}
\displaystyle =-2\sqrt{a}\sin a.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Evaluate }\lim \limits_{x\to\pi}\frac{\sqrt{5+\cos x}-2}{(\pi-x)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=\pi-x.
\displaystyle \text{As }x\to\pi,\ h\to0.
\displaystyle \lim \limits_{x\to\pi}\frac{\sqrt{5+\cos x}-2}{(\pi-x)^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{5+\cos(\pi-h)}-2}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{5-\cos h}-2}{h^2}
\displaystyle =\lim \limits_{h\to0}\left[\frac{\sqrt{5-\cos h}-2}{h^2}\times\frac{\sqrt{5-\cos h}+2}{\sqrt{5-\cos h}+2}\right]
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos h}{h^2\left(\sqrt{5-\cos h}+2\right)}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2\left(\frac{h}{2}\right)}{h^2\left(\sqrt{5-\cos h}+2\right)}
\displaystyle =\frac{1}{2}\lim \limits_{h\to0}\left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2
\displaystyle \qquad\times\lim \limits_{h\to0}\frac{1}{\sqrt{5-\cos h}+2}
\displaystyle =\frac{1}{2}\times1\times\frac{1}{\sqrt{5-1}+2}
\displaystyle =\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Evaluate }\lim \limits_{x\to a}\frac{\cos\sqrt{x}-\cos\sqrt{a}}{x-a},\quad a>0.
\displaystyle \text{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\left(\frac{\sqrt{x}+\sqrt{a}}{2}\right)\sin\left(\frac{\sqrt{x}-\sqrt{a}}{2}\right)}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}
\displaystyle =-\lim \limits_{x\to a}\frac{\sin\left(\frac{\sqrt{x}+\sqrt{a}}{2}\right)}{\sqrt{x}+\sqrt{a}}\frac{\sin\left(\frac{\sqrt{x}-\sqrt{a}}{2}\right)}{\frac{\sqrt{x}-\sqrt{a}}{2}}
\displaystyle =-\frac{\sin\left(\frac{\sqrt{a}+\sqrt{a}}{2}\right)}{\sqrt{a}+\sqrt{a}}\times1
\displaystyle =-\frac{\sin\sqrt{a}}{2\sqrt{a}}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Evaluate }\lim \limits_{x\to a}\frac{\sin\sqrt{x}-\sin\sqrt{a}}{x-a},\quad a>0.
\displaystyle \text{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\left(\frac{\sqrt{x}+\sqrt{a}}{2}\right)\sin\left(\frac{\sqrt{x}-\sqrt{a}}{2}\right)}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}
\displaystyle =\lim \limits_{x\to a}\frac{\cos\left(\frac{\sqrt{x}+\sqrt{a}}{2}\right)}{\sqrt{x}+\sqrt{a}}\frac{\sin\left(\frac{\sqrt{x}-\sqrt{a}}{2}\right)}{\frac{\sqrt{x}-\sqrt{a}}{2}}
\displaystyle =\frac{\cos\left(\frac{\sqrt{a}+\sqrt{a}}{2}\right)}{\sqrt{a}+\sqrt{a}}\times1
\displaystyle =\frac{\cos\sqrt{a}}{2\sqrt{a}}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Evaluate }\lim \limits_{x\to1}\frac{1-x^2}{\sin2\pi x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=1-h.
\displaystyle \text{As }x\to1,\ h\to0.
\displaystyle \lim \limits_{x\to1}\frac{1-x^2}{\sin2\pi x}
\displaystyle =\lim \limits_{h\to0}\frac{1-(1-h)^2}{\sin\left[2\pi(1-h)\right]}
\displaystyle =\lim \limits_{h\to0}\frac{2h-h^2}{-\sin2\pi h}
\displaystyle =\lim \limits_{h\to0}\frac{-(2-h)}{\frac{\sin2\pi h}{h}}
\displaystyle =\lim \limits_{h\to0}\frac{h-2}{2\pi\left(\frac{\sin2\pi h}{2\pi h}\right)}
\displaystyle =\frac{0-2}{2\pi\times1}
\displaystyle =-\frac{1}{\pi}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{f(x)-f\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}},\text{ where }f(x)=\sin2x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{4}+h.
\displaystyle \text{As }x\to\frac{\pi}{4},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{f(x)-f\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}
\displaystyle =\lim \limits_{h\to0}\frac{f\left(\frac{\pi}{4}+h\right)-f\left(\frac{\pi}{4}\right)}{h}
\displaystyle =\lim \limits_{h\to0}\frac{\sin\left(\frac{\pi}{2}+2h\right)-\sin\frac{\pi}{2}}{h}
\displaystyle =\lim \limits_{h\to0}\frac{\cos2h-1}{h}
\displaystyle =\lim \limits_{h\to0}\frac{-2\sin^2h}{h}
\displaystyle =-2\lim \limits_{h\to0}\left(\frac{\sin h}{h}\right)^2h
\displaystyle =-2\times1^2\times0=0.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Evaluate }\lim \limits_{x\to1}\frac{1+\cos\pi x}{(1-x)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=1-h.
\displaystyle \text{As }x\to1,\ h\to0.
\displaystyle \lim \limits_{x\to1}\frac{1+\cos\pi x}{(1-x)^2}
\displaystyle =\lim \limits_{h\to0}\frac{1+\cos\left[\pi(1-h)\right]}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos\pi h}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2\left(\frac{\pi h}{2}\right)}{h^2}
\displaystyle =\frac{\pi^2}{2}\lim \limits_{h\to0}\left(\frac{\sin\left(\frac{\pi h}{2}\right)}{\frac{\pi h}{2}}\right)^2
\displaystyle =\frac{\pi^2}{2}\times1
\displaystyle =\frac{\pi^2}{2}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Evaluate }\lim \limits_{x\to1}\frac{1-x^2}{\sin\pi x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=1-h.
\displaystyle \text{As }x\to1,\ h\to0.
\displaystyle \lim \limits_{x\to1}\frac{1-x^2}{\sin\pi x}
\displaystyle =\lim \limits_{h\to0}\frac{1-(1-h)^2}{\sin\left[\pi(1-h)\right]}
\displaystyle =\lim \limits_{h\to0}\frac{h(2-h)}{\sin\pi h}
\displaystyle =\lim \limits_{h\to0}\frac{2-h}{\pi\left(\frac{\sin\pi h}{\pi h}\right)}
\displaystyle =\frac{2-0}{\pi\times1}
\displaystyle =\frac{2}{\pi}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{1-\sin2x}{1+\cos4x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{4}-h.
\displaystyle \text{As }x\to\frac{\pi}{4},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{1-\sin2x}{1+\cos4x}
\displaystyle =\lim \limits_{h\to0}\frac{1-\sin\left(\frac{\pi}{2}-2h\right)}{1+\cos(\pi-4h)}
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos2h}{1-\cos4h}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2h}{2\sin^22h}
\displaystyle =\lim \limits_{h\to0}\frac{\sin^2h}{4\sin^2h\cos^2h}
\displaystyle =\lim \limits_{h\to0}\frac{1}{4\cos^2h}
\displaystyle =\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Evaluate }\lim \limits_{x\to\pi}\frac{1+\cos x}{\tan^2x}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\pi}\frac{1+\cos x}{\tan^2x}
\displaystyle =\lim \limits_{x\to\pi}\frac{(1+\cos x)\cos^2x}{\sin^2x}
\displaystyle =\lim \limits_{x\to\pi}\frac{(1+\cos x)\cos^2x}{1-\cos^2x}
\displaystyle =\lim \limits_{x\to\pi}\frac{(1+\cos x)\cos^2x}{(1-\cos x)(1+\cos x)}
\displaystyle =\lim \limits_{x\to\pi}\frac{\cos^2x}{1-\cos x}
\displaystyle =\frac{\cos^2\pi}{1-\cos\pi}
\displaystyle =\frac{(-1)^2}{1-(-1)}
\displaystyle =\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Evaluate }\lim \limits_{n\to\infty}n\sin\left(\frac{\pi}{4n}\right)\cos\left(\frac{\pi}{4n}\right).
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{n\to\infty}n\sin\left(\frac{\pi}{4n}\right)\cos\left(\frac{\pi}{4n}\right)
\displaystyle =\lim \limits_{n\to\infty}\left[\frac{\sin\left(\frac{\pi}{4n}\right)}{\frac{\pi}{4n}}\times\frac{\pi}{4}\times\cos\left(\frac{\pi}{4n}\right)\right]
\displaystyle =1\times\frac{\pi}{4}\times1
\displaystyle =\frac{\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Evaluate }\lim \limits_{n\to\infty}2^{n-1}\sin\left(\frac{a}{2^n}\right).
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{n\to\infty}2^{n-1}\sin\left(\frac{a}{2^n}\right)
\displaystyle =\lim \limits_{n\to\infty}\left[\frac{2^n}{2}\times\frac{\sin\left(\frac{a}{2^n}\right)}{\frac{a}{2^n}}\times\frac{a}{2^n}\right]
\displaystyle =\frac{a}{2}\lim \limits_{n\to\infty}\frac{\sin\left(\frac{a}{2^n}\right)}{\frac{a}{2^n}}
\displaystyle =\frac{a}{2}\times1
\displaystyle =\frac{a}{2}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Evaluate }\lim \limits_{n\to\infty}\frac{\sin\left(\frac{a}{2^n}\right)}{\sin\left(\frac{b}{2^n}\right)},\quad b\neq0.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{\sin\left(\frac{a}{2^n}\right)}{\sin\left(\frac{b}{2^n}\right)}
\displaystyle =\lim \limits_{n\to\infty}\left[\frac{\sin\left(\frac{a}{2^n}\right)}{\frac{a}{2^n}}\times\frac{\frac{b}{2^n}}{\sin\left(\frac{b}{2^n}\right)}\times\frac{a}{b}\right]
\displaystyle =1\times1\times\frac{a}{b}
\displaystyle =\frac{a}{b}.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Evaluate }\lim \limits_{x\to-1}\frac{x^2-x-2}{x^2+x+\sin(x+1)}.
\displaystyle \text{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{x-2}{x+\frac{\sin(x+1)}{x+1}}.
\displaystyle \text{Let }t=x+1,\text{ so that }x=t-1\text{ and }t\to0.
\displaystyle \frac{x-2}{x+\frac{\sin(x+1)}{x+1}}=\frac{t-3}{t-1+\frac{\sin t}{t}}.
\displaystyle t-1+\frac{\sin t}{t}=t+\left(\frac{\sin t}{t}-1\right).
\displaystyle \text{Also, }\frac{\sin t}{t}-1=\frac{\sin t-t}{t}=O(t^2).
\displaystyle \therefore t-1+\frac{\sin t}{t}\text{ has the same sign as }t\text{ near }0.
\displaystyle \lim \limits_{x\to-1^-}\frac{x^2-x-2}{x^2+x+\sin(x+1)}=+\infty.
\displaystyle \lim \limits_{x\to-1^+}\frac{x^2-x-2}{x^2+x+\sin(x+1)}=-\infty.
\displaystyle \text{Since the left-hand and right-hand limits are unequal, the limit does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Evaluate }\lim \limits_{x\to2}\frac{x^2-x-2}{x^2-2x+\sin(x-2)}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to2}\frac{x^2-x-2}{x^2-2x+\sin(x-2)}
\displaystyle =\lim \limits_{x\to2}\frac{(x-2)(x+1)}{x(x-2)+\sin(x-2)}
\displaystyle =\lim \limits_{x\to2}\frac{x+1}{x+\frac{\sin(x-2)}{x-2}}
\displaystyle =\frac{2+1}{2+1}
\displaystyle =1.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Evaluate }\lim \limits_{x\to1}(1-x)\tan\left(\frac{\pi x}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=1-h.
\displaystyle \text{As }x\to1,\ h\to0.
\displaystyle \lim \limits_{x\to1}(1-x)\tan\left(\frac{\pi x}{2}\right)
\displaystyle =\lim \limits_{h\to0}h\tan\left(\frac{\pi(1-h)}{2}\right)
\displaystyle =\lim \limits_{h\to0}h\tan\left(\frac{\pi}{2}-\frac{\pi h}{2}\right)
\displaystyle =\lim \limits_{h\to0}h\cot\left(\frac{\pi h}{2}\right)
\displaystyle =\lim \limits_{h\to0}\frac{h}{\tan\left(\frac{\pi h}{2}\right)}
\displaystyle =\frac{2}{\pi}\lim \limits_{h\to0}\frac{\frac{\pi h}{2}}{\tan\left(\frac{\pi h}{2}\right)}
\displaystyle =\frac{2}{\pi}\times1
\displaystyle =\frac{2}{\pi}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}.
\displaystyle \text{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)}{1-2\sin^2x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{\left(1-\frac{\sin x}{\cos x}\right)(1+\sqrt{2}\sin x)}{\cos2x}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{(\cos x-\sin x)(1+\sqrt{2}\sin x)}{\cos x(\cos^2x-\sin^2x)}
\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}\left(\frac{1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)}
\displaystyle =\frac{2}{1}
\displaystyle =2.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Evaluate }\lim \limits_{x\to\pi}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\pi-h.
\displaystyle \text{As }x\to\pi,\ h\to0.
\displaystyle \lim \limits_{x\to\pi}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2+\cos(\pi-h)}-1}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2-\cos h}-1}{h^2}
\displaystyle =\lim \limits_{h\to0}\frac{(\sqrt{2-\cos h}-1)(\sqrt{2-\cos h}+1)}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\lim \limits_{h\to0}\frac{1-\cos h}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\lim \limits_{h\to0}\frac{2\sin^2\left(\frac h2\right)}{h^2(\sqrt{2-\cos h}+1)}
\displaystyle =\frac{1}{2}\lim \limits_{h\to0}\frac{\sin^2\left(\frac h2\right)}{\left(\frac h2\right)^2}\times\frac{1}{\sqrt{2-\cos h}+1}
\displaystyle =\frac{1}{2}\times1\times\frac{1}{1+1}
\displaystyle =\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{\cos x}-\sqrt{\sin x}}{x-\frac{\pi}{4}}.
\displaystyle \text{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}}\frac{(\sqrt{\cos x}-\sqrt{\sin x})(\sqrt{\cos x}+\sqrt{\sin x})}{\left(x-\frac{\pi}{4}\right)(\sqrt{\cos x}+\sqrt{\sin x})}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{\left(x-\frac{\pi}{4}\right)(\sqrt{\cos x}+\sqrt{\sin x})}
\displaystyle \text{Let }x=\frac{\pi}{4}+h.
\displaystyle \text{As }x\to\frac{\pi}{4},\ h\to0.
\displaystyle =\lim \limits_{h\to0}\frac{\cos\left(\frac{\pi}{4}+h\right)-\sin\left(\frac{\pi}{4}+h\right)}{h\left(\sqrt{\cos\left(\frac{\pi}{4}+h\right)}+\sqrt{\sin\left(\frac{\pi}{4}+h\right)}\right)}
\displaystyle =\lim \limits_{h\to0}\frac{\frac{1}{\sqrt2}(\cos h-\sin h)-\frac{1}{\sqrt2}(\sin h+\cos h)}{h\left(\sqrt{\cos\left(\frac{\pi}{4}+h\right)}+\sqrt{\sin\left(\frac{\pi}{4}+h\right)}\right)}
\displaystyle =\lim \limits_{h\to0}\frac{-\sqrt2\,\sin h}{h\left(\sqrt{\cos\left(\frac{\pi}{4}+h\right)}+\sqrt{\sin\left(\frac{\pi}{4}+h\right)}\right)}
\displaystyle =-\sqrt2\lim \limits_{h\to0}\frac{\sin h}{h}\times\frac{1}{\sqrt{\cos\left(\frac{\pi}{4}+h\right)}+\sqrt{\sin\left(\frac{\pi}{4}+h\right)}}
\displaystyle =-\frac{\sqrt2}{2\left(\frac{1}{\sqrt2}\right)^{\frac14}}
\displaystyle =-\frac{1}{2^{\frac14}}=-2^{-\frac14}.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Evaluate }\lim \limits_{x\to1}\frac{1-\frac{1}{x}}{\sin\pi(x-1)}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to1}\frac{1-\frac{1}{x}}{\sin\pi(x-1)}
\displaystyle =\lim \limits_{x\to1}\frac{x-1}{x\sin\pi(x-1)}
\displaystyle \text{Let }h=x-1.
\displaystyle \text{As }x\to1,\ h\to0\text{ and }x=h+1.
\displaystyle =\lim \limits_{h\to0}\frac{h}{(h+1)\sin\pi h}
\displaystyle =\frac{1}{\pi}\lim \limits_{h\to0}\frac{\pi h}{\sin\pi h}\times\frac{1}{h+1}
\displaystyle =\frac{1}{\pi}\times1\times1
\displaystyle =\frac{1}{\pi}.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{6}}\frac{\cot^2x-3}{\mathrm{cosec}\,x-2}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{6}}\frac{\cot^2x-3}{\mathrm{cosec}\,x-2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{6}}\frac{\mathrm{cosec}^2x-1-3}{\mathrm{cosec}\,x-2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{6}}\frac{\mathrm{cosec}^2x-4}{\mathrm{cosec}\,x-2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{6}}\frac{(\mathrm{cosec}\,x-2)(\mathrm{cosec}\,x+2)}{\mathrm{cosec}\,x-2}
\displaystyle =\lim \limits_{x\to\frac{\pi}{6}}(\mathrm{cosec}\,x+2)
\displaystyle =2+2
\displaystyle =4.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{4}+h.
\displaystyle \text{As }x\to\frac{\pi}{4},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2}-\cos\left(\frac{\pi}{4}+h\right)-\sin\left(\frac{\pi}{4}+h\right)}{\left(4\left(\frac{\pi}{4}+h\right)-\pi\right)^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2}-\cos\left(\frac{\pi}{4}+h\right)-\sin\left(\frac{\pi}{4}+h\right)}{16h^2}
\displaystyle =\lim \limits_{h\to0}\frac{\sqrt{2}-\sqrt{2}\cos h}{16h^2}
\displaystyle =\frac{\sqrt{2}}{16}\lim \limits_{h\to0}\frac{1-\cos h}{h^2}
\displaystyle =\frac{\sqrt{2}}{16}\lim \limits_{h\to0}\frac{2\sin^2\left(\frac h2\right)}{h^2}
\displaystyle =\frac{\sqrt{2}}{32}\lim \limits_{h\to0}\frac{\sin^2\left(\frac h2\right)}{\left(\frac h2\right)^2}
\displaystyle =\frac{\sqrt{2}}{32}\times1
\displaystyle =\frac{1}{16\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)\sin x-2\cos x}{\left(\frac{\pi}{2}-x\right)+\cot x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\frac{\pi}{2}-h.
\displaystyle \text{As }x\to\frac{\pi}{2},\ h\to0.
\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)\sin x-2\cos x}{\left(\frac{\pi}{2}-x\right)+\cot x}
\displaystyle =\lim \limits_{h\to0}\frac{h\sin\left(\frac{\pi}{2}-h\right)-2\cos\left(\frac{\pi}{2}-h\right)}{h+\cot\left(\frac{\pi}{2}-h\right)}
\displaystyle =\lim \limits_{h\to0}\frac{h\cos h-2\sin h}{h+\tan h}
\displaystyle =\lim \limits_{h\to0}\frac{\cos h-2\frac{\sin h}{h}}{1+\frac{\tan h}{h}}
\displaystyle =\frac{1-2(1)}{1+1}
\displaystyle =-\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Evaluate }\lim \limits_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{\left(\frac{\pi}{4}-x\right)(\cos x+\sin x)}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{\left(\frac{\pi}{4}-x\right)(\cos x+\sin x)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\left(\sin\frac{\pi}{4}\cos x-\cos\frac{\pi}{4}\sin x\right)}{\left(\frac{\pi}{4}-x\right)(\cos x+\sin x)}
\displaystyle =\lim \limits_{x\to\frac{\pi}{4}}\left(\frac{\sqrt{2}}{\cos x+\sin x}\times\frac{\sin\left(\frac{\pi}{4}-x\right)}{\frac{\pi}{4}-x}\right)
\displaystyle =\frac{\sqrt{2}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}\times1
\displaystyle =1.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Evaluate }\lim \limits_{x\to\pi}\frac{1-\sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to\pi}\frac{1-\sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)}
\displaystyle =\lim \limits_{x\to\pi}\frac{\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)^2}{\left(\cos^2\frac{x}{4}-\sin^2\frac{x}{4}\right)\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)}
\displaystyle =\lim \limits_{x\to\pi}\frac{\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)^2}{\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)^2\left(\cos\frac{x}{4}+\sin\frac{x}{4}\right)}
\displaystyle =\lim \limits_{x\to\pi}\frac{1}{\cos\frac{x}{4}+\sin\frac{x}{4}}
\displaystyle =\frac{1}{\cos\frac{\pi}{4}+\sin\frac{\pi}{4}}
\displaystyle =\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}
\displaystyle =\frac{1}{\sqrt{2}}.
\displaystyle \\

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