\displaystyle \textbf{Question 1. }\lim_{n\to\infty}\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{3}\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}
\displaystyle \therefore \lim_{n\to\infty}\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}
\displaystyle =\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}
\displaystyle =\frac{1}{6}\lim_{n\to\infty}\frac{(n+1)(2n+1)}{n^2}
\displaystyle =\frac{1}{6}\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)
\displaystyle =\frac{1}{6}(1)(2)
\displaystyle =\frac{1}{3}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 2. }\lim_{x\to0}\frac{\sin2x}{x}\text{ is equal to}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin2x}{x}
\displaystyle =2\lim_{x\to0}\frac{\sin2x}{2x}
\displaystyle =2(1)
\displaystyle =2
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 3. }\text{If }f(x)=x\sin\left(\frac{1}{x}\right),\ x\neq0,\text{ then }\lim_{x\to0}f(x)=
\displaystyle \text{(a) }1\qquad \text{(b) }0\qquad \text{(c) }-1\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin\left(\frac{1}{x}\right)\leq1
\displaystyle \Rightarrow -x\leq x\sin\left(\frac{1}{x}\right)\leq x
\displaystyle \text{As }x\to0,\ -x\to0\text{ and }x\to0
\displaystyle \therefore \lim_{x\to0}x\sin\left(\frac{1}{x}\right)=0
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 4. }\lim_{x\to0}\frac{1-\cos2x}{x}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) }4
\displaystyle \text{Answer:}
\displaystyle 1-\cos2x=2\sin^2x
\displaystyle \therefore \lim_{x\to0}\frac{1-\cos2x}{x}
\displaystyle =\lim_{x\to0}\frac{2\sin^2x}{x}
\displaystyle =2\lim_{x\to0}\left(\frac{\sin x}{x}\right)\sin x
\displaystyle =2(1)(0)
\displaystyle =0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 5. }\lim_{x\to0}\frac{(1-\cos2x)\sin5x}{x^2\sin3x}\text{ is equal to}
\displaystyle \text{(a) }\frac{10}{3}\qquad \text{(b) }\frac{3}{10}\qquad \text{(c) }\frac{6}{5}\qquad \text{(d) }\frac{5}{6}
\displaystyle \text{Answer:}
\displaystyle 1-\cos2x=2\sin^2x
\displaystyle \therefore \lim_{x\to0}\frac{(1-\cos2x)\sin5x}{x^2\sin3x}
\displaystyle =\lim_{x\to0}\frac{2\sin^2x\sin5x}{x^2\sin3x}
\displaystyle =2\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\left(\frac{\sin5x}{\sin3x}\right)
\displaystyle =2(1)^2\cdot\frac{5}{3}
\displaystyle =\frac{10}{3}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 6. }\lim_{x\to0}\frac{x}{\tan x}\text{ is equal to}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }4\qquad \text{(d) not defined}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{x}{\tan x}
\displaystyle =\lim_{x\to0}\frac{x\cos x}{\sin x}
\displaystyle =\lim_{x\to0}\frac{x}{\sin x}\cdot\lim_{x\to0}\cos x
\displaystyle =1\cdot1
\displaystyle =1
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 7. }\lim_{n\to\infty}\left\{\frac{1}{1-n^2}+\frac{2}{1-n^2}+\cdots+\frac{n}{1-n^2}\right\}\text{ is equal to}
\displaystyle \text{(a) }0\qquad \text{(b) }-\frac{1}{2}\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{1-n^2}+\frac{2}{1-n^2}+\cdots+\frac{n}{1-n^2}
\displaystyle =\frac{1+2+\cdots+n}{1-n^2}
\displaystyle =\frac{\frac{n(n+1)}{2}}{1-n^2}
\displaystyle =\frac{n(n+1)}{2(1-n^2)}
\displaystyle =\frac{n(n+1)}{2(1-n)(1+n)}
\displaystyle =\frac{n}{2(1-n)}
\displaystyle \therefore \lim_{n\to\infty}\frac{n}{2(1-n)}=-\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 8. }\lim_{x\to\infty}\frac{\sin x}{x}\text{ equals}
\displaystyle \text{(a) }0\qquad \text{(b) }\infty\qquad \text{(c) }1\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin x\leq1
\displaystyle \Rightarrow -\frac{1}{x}\leq\frac{\sin x}{x}\leq\frac{1}{x}
\displaystyle \text{As }x\to\infty,\ -\frac{1}{x}\to0\text{ and }\frac{1}{x}\to0
\displaystyle \therefore \lim_{x\to\infty}\frac{\sin x}{x}=0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 9. }\lim_{x\to0}\frac{\sin x^\circ}{x}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }\pi\qquad \text{(c) }x\qquad \text{(d) }\frac{\pi}{180}
\displaystyle \text{Answer:}
\displaystyle \sin x^\circ=\sin\left(\frac{\pi x}{180}\right)
\displaystyle \therefore \lim_{x\to0}\frac{\sin x^\circ}{x}
\displaystyle =\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x}
\displaystyle =\frac{\pi}{180}\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}
\displaystyle =\frac{\pi}{180}(1)
\displaystyle =\frac{\pi}{180}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 10. }\lim_{x\to3}\frac{x-3}{|x-3|}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }0\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{When }x\to3^+,\ x-3>0
\displaystyle \therefore \frac{x-3}{|x-3|}=1
\displaystyle \Rightarrow \text{R.H.L.}=1
\displaystyle \text{When }x\to3^-,\ x-3<0
\displaystyle \therefore \frac{x-3}{|x-3|}=-1
\displaystyle \Rightarrow \text{L.H.L.}=-1
\displaystyle \text{Since L.H.L.}\neq\text{R.H.L., the limit does not exist.}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 11. }\lim_{x\to a}\frac{x^n-a^n}{x-a}\text{ is equal to}
\displaystyle \text{(a) }na^n\qquad \text{(b) }na^{n-1}\qquad \text{(c) }na\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1})
\displaystyle \therefore \frac{x^n-a^n}{x-a}=x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}
\displaystyle \therefore \lim_{x\to a}\frac{x^n-a^n}{x-a}=a^{n-1}+a^{n-1}+\cdots+a^{n-1}
\displaystyle =na^{n-1}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 12. }\lim_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\cos x-1}{\cot x-1}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{\sqrt{2}}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{2\sqrt{2}}\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\cos x-1}{\cot x-1}
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\cos x-1}{\frac{\cos x}{\sin x}-1}
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{\sin x(\sqrt{2}\cos x-1)}{\cos x-\sin x}
\displaystyle \text{Using L'Hospital's rule,}
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{\cos x(\sqrt{2}\cos x-1)-\sqrt{2}\sin^2x}{-\sin x-\cos x}
\displaystyle =\frac{\frac{1}{\sqrt{2}}(1-1)-\sqrt{2}\cdot\frac{1}{2}}{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}
\displaystyle =\frac{-\frac{1}{\sqrt{2}}}{-\sqrt{2}}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 13. }\lim_{x\to\infty}\frac{\sqrt{x^2-1}}{2x+1}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }0\qquad \text{(c) }-1\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\infty}\frac{\sqrt{x^2-1}}{2x+1}
\displaystyle =\lim_{x\to\infty}\frac{x\sqrt{1-\frac{1}{x^2}}}{x\left(2+\frac{1}{x}\right)}
\displaystyle =\lim_{x\to\infty}\frac{\sqrt{1-\frac{1}{x^2}}}{2+\frac{1}{x}}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 14. }\lim_{h\to0}2\left\{\frac{\sqrt{3}\sin\left(\frac{\pi}{6}+h\right)-\cos\left(\frac{\pi}{6}+h\right)}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}\right\}\text{ is equal to}
\displaystyle \text{(a) }\frac{2}{3}\qquad \text{(b) }\frac{4}{3}\qquad \text{(c) }-2\sqrt{3}\qquad \text{(d) }-\frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle \sqrt{3}\sin\left(\frac{\pi}{6}+h\right)-\cos\left(\frac{\pi}{6}+h\right)
\displaystyle =\sqrt{3}\left(\frac{1}{2}\cos h+\frac{\sqrt{3}}{2}\sin h\right)-\left(\frac{\sqrt{3}}{2}\cos h-\frac{1}{2}\sin h\right)
\displaystyle =2\sin h
\displaystyle \therefore \lim_{h\to0}2\left\{\frac{2\sin h}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}\right\}
\displaystyle =\lim_{h\to0}\frac{4\sin h}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}
\displaystyle =\frac{4}{\sqrt{3}}\lim_{h\to0}\frac{\sin h}{h}\cdot\frac{1}{\sqrt{3}\cos h-\sin h}
\displaystyle =\frac{4}{\sqrt{3}}\cdot1\cdot\frac{1}{\sqrt{3}}
\displaystyle =\frac{4}{3}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 15. }\lim_{h\to0}\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}\text{ is equal to}
\displaystyle \text{(a) }-\frac{1}{12}\qquad \text{(b) }-\frac{4}{3}\qquad \text{(c) }-\frac{16}{3}\qquad \text{(d) }-\frac{1}{48}
\displaystyle \text{Answer:}
\displaystyle \lim_{h\to0}\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}
\displaystyle =\lim_{h\to0}\frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}
\displaystyle \text{Let }a=\sqrt[3]{8+h}
\displaystyle \therefore a^3=8+h
\displaystyle \Rightarrow h=a^3-8=(a-2)(a^2+2a+4)
\displaystyle \therefore \lim_{h\to0}\frac{2-a}{2a(a-2)(a^2+2a+4)}
\displaystyle =\lim_{a\to2}\frac{-(a-2)}{2a(a-2)(a^2+2a+4)}
\displaystyle =\lim_{a\to2}\frac{-1}{2a(a^2+2a+4)}
\displaystyle =\frac{-1}{2(2)(4+4+4)}
\displaystyle =-\frac{1}{48}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 16. }\lim_{n\to\infty}\left\{\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\cdots+\frac{1}{(2n+1)(2n+3)}\right\}\text{ is equal to}
\displaystyle \text{(a) }0\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{9}\qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)
\displaystyle \therefore \frac{1}{1.3}+\frac{1}{3.5}+\cdots+\frac{1}{(2n+1)(2n+3)}
\displaystyle =\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots+\frac{1}{2n+1}-\frac{1}{2n+3}\right)
\displaystyle =\frac{1}{2}\left(1-\frac{1}{2n+3}\right)
\displaystyle \therefore \lim_{n\to\infty}\frac{1}{2}\left(1-\frac{1}{2n+3}\right)=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 17. }\lim_{x\to1}\frac{\sin\pi x}{x-1}\text{ is equal to}
\displaystyle \text{(a) }-\pi\qquad \text{(b) }\pi\qquad \text{(c) }-\frac{1}{\pi}\qquad \text{(d) }\frac{1}{\pi}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{\sin\pi x}{x-1}
\displaystyle =\lim_{x\to1}\frac{\sin(\pi+\pi(x-1))}{x-1}
\displaystyle =\lim_{x\to1}\frac{-\sin(\pi(x-1))}{x-1}
\displaystyle =-\pi\lim_{x\to1}\frac{\sin(\pi(x-1))}{\pi(x-1)}
\displaystyle =-\pi
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 18. }\text{If }\lim_{x\to1}\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}=5050,\text{ then }n\text{ equals}
\displaystyle \text{(a) }10\qquad \text{(b) }100\qquad \text{(c) }150\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=x+x^2+x^3+\cdots+x^n
\displaystyle \lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1)
\displaystyle f'(x)=1+2x+3x^2+\cdots+nx^{n-1}
\displaystyle \therefore f'(1)=1+2+3+\cdots+n=\frac{n(n+1)}{2}
\displaystyle \frac{n(n+1)}{2}=5050
\displaystyle \Rightarrow n(n+1)=10100
\displaystyle \Rightarrow n=100
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 19. }\text{The value of }\lim_{x\to\infty}\frac{\sqrt{1+x^4}+(1+x^2)}{x^2}\text{ is}
\displaystyle \text{(a) }-1\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\infty}\frac{\sqrt{1+x^4}+(1+x^2)}{x^2}
\displaystyle =\lim_{x\to\infty}\left(\sqrt{\frac{1}{x^4}+1}+\frac{1}{x^2}+1\right)
\displaystyle =1+0+1
\displaystyle =2
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 20. }\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }2\qquad \text{(c) }0\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sqrt{1+x}-1}{x}
\displaystyle =\lim_{x\to0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}
\displaystyle =\lim_{x\to0}\frac{1+x-1}{x(\sqrt{1+x}+1)}
\displaystyle =\lim_{x\to0}\frac{1}{\sqrt{1+x}+1}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 21. }\lim_{x\to\frac{\pi}{3}}\frac{\sin\left(\frac{\pi}{3}-x\right)}{2\cos x-1}\text{ is equal to}
\displaystyle \text{(a) }\sqrt{3}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{\sqrt{3}}\qquad \text{(d) }\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=x-\frac{\pi}{3}
\displaystyle \text{Then, as }x\to\frac{\pi}{3},\ h\to0
\displaystyle \frac{\sin\left(\frac{\pi}{3}-x\right)}{2\cos x-1} =\frac{\sin(-h)}{2\cos\left(\frac{\pi}{3}+h\right)-1}
\displaystyle =\frac{-\sin h}{2\left(\frac{1}{2}\cos h-\frac{\sqrt{3}}{2}\sin h\right)-1}
\displaystyle =\frac{-\sin h}{\cos h-\sqrt{3}\sin h-1}
\displaystyle =\frac{-\sin h}{(\cos h-1)-\sqrt{3}\sin h}
\displaystyle =\frac{-1}{\frac{\cos h-1}{\sin h}-\sqrt{3}}
\displaystyle =\frac{-1}{0-\sqrt{3}}=\frac{1}{\sqrt{3}}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 22. }\lim_{x\to3}\frac{\sum_{r=1}^{n}x^r-\sum_{r=1}^{n}3^r}{x-3}\text{ is equal to}
\displaystyle \text{(a) }\frac{(2n-1)3^n}{4}\qquad \text{(b) }\frac{(2n-1)3^n+1}{4}\qquad \text{(c) }(2n-1)3^n+1\qquad \text{(d) }\frac{(2n-1)3^n-1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\sum_{r=1}^{n}x^r
\displaystyle \lim_{x\to3}\frac{f(x)-f(3)}{x-3}=f'(3)
\displaystyle f'(x)=\sum_{r=1}^{n}rx^{r-1}
\displaystyle \therefore f'(3)=1+2\cdot3+3\cdot3^2+\cdots+n3^{n-1}
\displaystyle =\frac{(2n-1)3^n+1}{4}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 23. }\lim_{n\to\infty}\frac{1-2+3-4+5-6+\cdots+(2n-1)-2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }-\frac{1}{2}\qquad \text{(c) }1\qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle 1-2+3-4+\cdots+(2n-1)-2n
\displaystyle =(1-2)+(3-4)+\cdots+((2n-1)-2n)
\displaystyle =-1-1-\cdots-1=-n
\displaystyle \therefore \lim_{n\to\infty}\frac{-n}{\sqrt{n^2+1}+\sqrt{n^2-1}}
\displaystyle =\lim_{n\to\infty}\frac{-n}{n\sqrt{1+\frac{1}{n^2}}+n\sqrt{1-\frac{1}{n^2}}}
\displaystyle =\lim_{n\to\infty}\frac{-1}{\sqrt{1+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n^2}}}
\displaystyle =-\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 24. }\text{If }f(x)=\begin{cases}x\sin\frac{1}{x},&x\neq0\\0,&x=0\end{cases}\text{, then }\lim_{x\to0}f(x)\text{ equals}
\displaystyle \text{(a) }1\qquad \text{(b) }0\qquad \text{(c) }-1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin\frac{1}{x}\leq1
\displaystyle \Rightarrow -|x|\leq x\sin\frac{1}{x}\leq |x|
\displaystyle \text{As }x\to0,\ -|x|\to0\text{ and }|x|\to0
\displaystyle \therefore \lim_{x\to0}x\sin\frac{1}{x}=0
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 25. }\lim_{n\to\infty}\frac{n!}{(n+1)!+n!}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }0\qquad \text{(c) }2\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \frac{n!}{(n+1)!+n!}=\frac{n!}{(n+1)n!+n!}
\displaystyle =\frac{n!}{n!(n+1+1)}
\displaystyle =\frac{1}{n+2}
\displaystyle \therefore \lim_{n\to\infty}\frac{1}{n+2}=0
\displaystyle \text{Hence, the correct option is (b).}
\\

\displaystyle \textbf{Question 26. }\lim_{x\to\frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin2x}\text{ is equal to}
\displaystyle \text{(a) }5\sqrt{2}\qquad \text{(b) }3\sqrt{2}\qquad \text{(c) }\sqrt{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (\cos x+\sin x)^2=1+\sin2x
\displaystyle \therefore 1-\sin2x=2-(\cos x+\sin x)^2
\displaystyle \text{Let }y=\cos x+\sin x
\displaystyle \text{As }x\to\frac{\pi}{4},\ y\to\sqrt{2}
\displaystyle \lim_{x\to\frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin2x}
\displaystyle =\lim_{y\to\sqrt{2}}\frac{4\sqrt{2}-y^5}{2-y^2}
\displaystyle =\lim_{y\to\sqrt{2}}\frac{(\sqrt{2})^5-y^5}{(\sqrt{2})^2-y^2}
\displaystyle =\frac{5(\sqrt{2})^4}{2\sqrt{2}}
\displaystyle =\frac{5\cdot4}{2\sqrt{2}}
\displaystyle =5\sqrt{2}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 27. }\lim_{x\to2}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{8\sqrt{3}}\qquad \text{(b) }\frac{1}{\sqrt{3}}\qquad \text{(c) }8\sqrt{3}\qquad \text{(d) }\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\sqrt{1+\sqrt{2+x}}
\displaystyle \lim_{x\to2}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}=f'(2)
\displaystyle f'(x)=\frac{1}{2\sqrt{1+\sqrt{2+x}}}\cdot\frac{1}{2\sqrt{2+x}}
\displaystyle \therefore f'(2)=\frac{1}{2\sqrt{3}}\cdot\frac{1}{4}
\displaystyle =\frac{1}{8\sqrt{3}}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 28. }\lim_{x\to\infty}a^x\sin\left(\frac{b}{a^x}\right),\ a,b>1\text{ is equal to}
\displaystyle \text{(a) }b\qquad \text{(b) }a\qquad \text{(c) }a\log_e b\qquad \text{(d) }b\log_e a
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\infty}a^x\sin\left(\frac{b}{a^x}\right)
\displaystyle =b\lim_{x\to\infty}\frac{\sin\left(\frac{b}{a^x}\right)}{\frac{b}{a^x}}
\displaystyle \text{As }x\to\infty,\ \frac{b}{a^x}\to0
\displaystyle \therefore \lim_{x\to\infty}a^x\sin\left(\frac{b}{a^x}\right)=b
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 29. }\lim_{x\to0}\frac{8}{x^8}\left\{1-\cos\frac{x^2}{2}-\cos\frac{x^2}{4}+\cos\frac{x^2}{2}\cos\frac{x^2}{4}\right\}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{16}\qquad \text{(b) }-\frac{1}{16}\qquad \text{(c) }\frac{1}{32}\qquad \text{(d) }-\frac{1}{32}
\displaystyle \text{Answer:}
\displaystyle 1-\cos\frac{x^2}{2}-\cos\frac{x^2}{4}+\cos\frac{x^2}{2}\cos\frac{x^2}{4}
\displaystyle =\left(1-\cos\frac{x^2}{2}\right)\left(1-\cos\frac{x^2}{4}\right)
\displaystyle \therefore \lim_{x\to0}\frac{8}{x^8}\left(1-\cos\frac{x^2}{2}\right)\left(1-\cos\frac{x^2}{4}\right)
\displaystyle =8\lim_{x\to0}\frac{1-\cos\frac{x^2}{2}}{x^4}\cdot\frac{1-\cos\frac{x^2}{4}}{x^4}
\displaystyle =8\cdot\frac{1}{8}\cdot\frac{1}{32}
\displaystyle =\frac{1}{32}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 30. }\text{If }\alpha\text{ is a repeated root of }ax^2+bx+c=0,\text{ then }\lim_{x\to\alpha}\frac{\tan(ax^2+bx+c)}{(x-\alpha)^2}\text{ is}
\displaystyle \text{(a) }a\qquad \text{(b) }b\qquad \text{(c) }c\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Since }\alpha\text{ is a repeated root,}
\displaystyle ax^2+bx+c=a(x-\alpha)^2
\displaystyle \therefore \lim_{x\to\alpha}\frac{\tan(ax^2+bx+c)}{(x-\alpha)^2}
\displaystyle =\lim_{x\to\alpha}\frac{\tan(a(x-\alpha)^2)}{(x-\alpha)^2}
\displaystyle =a\lim_{x\to\alpha}\frac{\tan(a(x-\alpha)^2)}{a(x-\alpha)^2}
\displaystyle =a
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 31. }\text{The value of }\lim_{x\to0}\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}}\text{ is}
\displaystyle \text{(a) }a\qquad \text{(b) }\sqrt{a}\qquad \text{(c) }-a\qquad \text{(d) }-\sqrt{a}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}}
\displaystyle =\frac{-\frac{a}{2a}-\frac{a}{2a}}{\frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{a}}}
\displaystyle =\frac{-1}{\frac{1}{\sqrt{a}}}
\displaystyle =-\sqrt{a}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 32. }\text{The value of }\lim_{x\to0}\frac{1-\cos x+2\sin x-\sin^3x-x^2+3x^4}{\tan^3x-6\sin^2x+x-5x^3}\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }-1\qquad \text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \text{Putting }x=0\text{ gives }\frac{0}{0}\text{ form.}
\displaystyle \text{Using L'Hospital's rule,}
\displaystyle =\lim_{x\to0}\frac{\sin x+2\cos x-3\sin^2x\cos x-2x+12x^3}{3\tan^2x\sec^2x-12\sin x\cos x+1-15x^2}
\displaystyle =\frac{0+2-0-0+0}{0-0+1-0}
\displaystyle =2
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 33. }\lim_{\theta\to\frac{\pi}{2}}\frac{1-\sin\theta}{\left(\frac{\pi}{2}-\theta\right)\cos\theta}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) }-\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Let }h=\frac{\pi}{2}-\theta
\displaystyle \text{Then, as }\theta\to\frac{\pi}{2},\ h\to0
\displaystyle \frac{1-\sin\theta}{\left(\frac{\pi}{2}-\theta\right)\cos\theta}=\frac{1-\cos h}{h\sin h}
\displaystyle =\frac{1-\cos h}{h^2}\cdot\frac{h}{\sin h}
\displaystyle =\frac{1}{2}\cdot1
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 34. }\text{The value of }\lim_{x\to\frac{\pi}{2}}(\sec x-\tan x)\text{ is}
\displaystyle \text{(a) }2\qquad \text{(b) }-1\qquad \text{(c) }1\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sec x-\tan x=\frac{1}{\cos x}-\frac{\sin x}{\cos x}
\displaystyle =\frac{1-\sin x}{\cos x}
\displaystyle =\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}
\displaystyle =\frac{1-\sin^2x}{\cos x(1+\sin x)}
\displaystyle =\frac{\cos^2x}{\cos x(1+\sin x)}
\displaystyle =\frac{\cos x}{1+\sin x}
\displaystyle \therefore \lim_{x\to\frac{\pi}{2}}(\sec x-\tan x)=\frac{0}{1+1}=0
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 35. }\text{The value of }\lim_{n\to\infty}\frac{n!}{(n+1)!-n!}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }-1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{n!}{(n+1)!-n!}=\frac{n!}{(n+1)n!-n!}
\displaystyle =\frac{n!}{n\cdot n!}
\displaystyle =\frac{1}{n}
\displaystyle \therefore \lim_{n\to\infty}\frac{1}{n}=0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 36. }\text{The value of }\lim_{n\to\infty}\frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }-1\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}=\frac{(n+2)(n+1)!+(n+1)!}{(n+2)(n+1)!-(n+1)!}
\displaystyle =\frac{(n+3)(n+1)!}{(n+1)(n+1)!}
\displaystyle =\frac{n+3}{n+1}
\displaystyle \therefore \lim_{n\to\infty}\frac{n+3}{n+1}=1
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 37. }\text{The value of }\lim_{x\to\infty}\frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}}\text{ is}
\displaystyle \text{(a) }10\qquad \text{(b) }100\qquad \text{(c) }10^{10}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}}
\displaystyle =\frac{x^{10}\left(1+\frac{1}{x}\right)^{10}+x^{10}\left(1+\frac{2}{x}\right)^{10}+\cdots+x^{10}\left(1+\frac{100}{x}\right)^{10}}{x^{10}\left(1+\frac{10^{10}}{x^{10}}\right)}
\displaystyle =\frac{\left(1+\frac{1}{x}\right)^{10}+\left(1+\frac{2}{x}\right)^{10}+\cdots+\left(1+\frac{100}{x}\right)^{10}}{1+\frac{10^{10}}{x^{10}}}
\displaystyle \therefore \lim_{x\to\infty}\frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}}{x^{10}+10^{10}}
\displaystyle =\frac{1+1+\cdots+1}{1}
\displaystyle =100
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 38. }\text{The value of }\lim_{n\to\infty}\left\{\frac{1+2+3+\cdots+n}{n+2}-\frac{n}{2}\right\}\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }1\qquad \text{(c) }-1\qquad \text{(d) }-\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{1+2+3+\cdots+n}{n+2}-\frac{n}{2}
\displaystyle =\frac{\frac{n(n+1)}{2}}{n+2}-\frac{n}{2}
\displaystyle =\frac{n(n+1)}{2(n+2)}-\frac{n}{2}
\displaystyle =\frac{n(n+1)-n(n+2)}{2(n+2)}
\displaystyle =\frac{-n}{2(n+2)}
\displaystyle \therefore \lim_{n\to\infty}\frac{-n}{2(n+2)}=-\frac{1}{2}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 39. }\lim_{x\to1}[x-1]\text{, where }[\cdot]\text{ is the greatest integer function, is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }0\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to1^+,\ x-1\to0^+,\text{ so }[x-1]=0
\displaystyle \therefore \text{R.H.L.}=0
\displaystyle \text{As }x\to1^-,\ x-1\to0^-,\text{ so }[x-1]=-1
\displaystyle \therefore \text{L.H.L.}=-1
\displaystyle \text{Since L.H.L.}\neq\text{R.H.L., the limit does not exist.}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 40. }\lim_{x\to\infty}\frac{|x|}{x}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }0\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to\infty,\ x>0
\displaystyle \therefore |x|=x
\displaystyle \therefore \frac{|x|}{x}=\frac{x}{x}=1
\displaystyle \therefore \lim_{x\to\infty}\frac{|x|}{x}=1
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 41. }\lim_{x\to0}\frac{|\sin x|}{x}\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }0\qquad \text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^+,\ \frac{|\sin x|}{x}=\frac{\sin x}{x}
\displaystyle \therefore \text{R.H.L.}=1
\displaystyle \text{As }x\to0^-,\ \frac{|\sin x|}{x}=\frac{-\sin x}{x}
\displaystyle \therefore \text{L.H.L.}=-1
\displaystyle \text{Since L.H.L.}\neq\text{R.H.L., the limit does not exist.}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 42. }\text{If }f(x)=\begin{cases}\frac{\sin[x]}{[x]},&[x]\neq0\\0,&[x]=0\end{cases}\text{, where }[\cdot]\text{ denotes the greatest integer function, then} \\ \lim_{x\to0}f(x)\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }0\qquad \text{(c) }-1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^+,\ [x]=0
\displaystyle \therefore f(x)=0
\displaystyle \therefore \text{R.H.L.}=0
\displaystyle \text{As }x\to0^-,\ [x]=-1
\displaystyle \therefore f(x)=\frac{\sin(-1)}{-1}=\sin1
\displaystyle \therefore \text{L.H.L.}=\sin1
\displaystyle \text{Since L.H.L.}\neq\text{R.H.L., the limit does not exist.}
\displaystyle \therefore \text{Correct option is (d) none of these}
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