Class 11: Limits – Very Short Answer Questions

$\displaystyle \textbf{Question 1. }\text{Write the value of }\lim_{x\to0}\frac{\sqrt{1-\cos 2x}}{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \lim_{x\to0}\frac{\sqrt{1-\cos 2x}}{x}$
$\displaystyle 1-\cos 2x=2\sin^2x$
$\displaystyle \therefore \frac{\sqrt{1-\cos 2x}}{x}=\frac{\sqrt{2\sin^2x}}{x}$
$\displaystyle =\frac{\sqrt{2}\,|\sin x|}{x}$
$\displaystyle \text{Right hand limit }=\sqrt{2}$
$\displaystyle \text{Left hand limit }=-\sqrt{2}$
$\displaystyle \text{Since L.H.L.}\neq\text{R.H.L., the limit does not exist.}$
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$\displaystyle \textbf{Question 2. }\text{Write the value of }\lim_{x\to0^-}[x].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{As }x\to0^-,\ x\text{ is negative and very close to }0$
$\displaystyle \therefore [x]=-1$
$\displaystyle \therefore \lim_{x\to0^-}[x]=-1$
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$\displaystyle \textbf{Question 3. }\text{Write the value of }\lim_{x\to0^+}[x].$
$\displaystyle \text{Answer:}$
$\displaystyle [x]\text{ denotes the greatest integer function.}$
$\displaystyle \text{As }x\to0^+,\text{ we have }0<x<1$
$\displaystyle \therefore [x]=0$
$\displaystyle \therefore \lim_{x\to0^+}[x]=0$
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$\displaystyle \textbf{Question 4. }\text{Write the value of }\lim_{x\to1^-}(x-[x]).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{As }x\to1^-,\ [x]=0$
$\displaystyle \therefore x-[x]=x$
$\displaystyle \therefore \lim_{x\to1^-}(x-[x])=1$
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$\displaystyle \textbf{Question 5. }\text{Write the value of }\lim_{x\to0^-}\frac{\sin[x]}{[x]}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{As }x\to0^-,\ [x]=-1$
$\displaystyle \therefore \frac{\sin[x]}{[x]}=\frac{\sin(-1)}{-1}$
$\displaystyle =\sin1$
$\displaystyle \therefore \lim_{x\to0^-}\frac{\sin[x]}{[x]}=\sin1$
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$\displaystyle \textbf{Question 6. }\text{Write the value of }\lim_{x\to\pi}\frac{\sin x}{x-\pi}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x-\pi=t$
$\displaystyle \text{Then }x=t+\pi$
$\displaystyle \therefore \frac{\sin x}{x-\pi}=\frac{\sin(t+\pi)}{t}$
$\displaystyle =\frac{-\sin t}{t}$
$\displaystyle \therefore \lim_{x\to\pi}\frac{\sin x}{x-\pi}=-1$
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$\displaystyle \textbf{Question 7. }\text{Write the value of }\lim_{x\to\infty}\frac{\sin x}{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq\sin x\leq1$
$\displaystyle \therefore -\frac{1}{x}\leq\frac{\sin x}{x}\leq\frac{1}{x}$
$\displaystyle \text{As }x\to\infty,\ \frac{1}{x}\to0$
$\displaystyle \therefore \lim_{x\to\infty}\frac{\sin x}{x}=0$
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$\displaystyle \textbf{Question 8. }\text{Write the value of }\lim_{x\to2}\frac{|x-2|}{x-2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x>2,\ \frac{|x-2|}{x-2}=1$
$\displaystyle \text{For }x<2,\ \frac{|x-2|}{x-2}=-1$
$\displaystyle \text{L.H.L.}=-1,\qquad \text{R.H.L.}=1$
$\displaystyle \therefore \lim_{x\to2}\frac{|x-2|}{x-2}\text{ does not exist}$
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$\displaystyle \textbf{Question 9. }\text{Write the value of }\lim_{x\to0}\frac{\sin x^\circ}{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle x^\circ=\frac{\pi x}{180}\text{ radians}$
$\displaystyle \therefore \sin x^\circ=\sin\left(\frac{\pi x}{180}\right)$
$\displaystyle \lim_{x\to0}\frac{\sin x^\circ}{x}=\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x}$
$\displaystyle =\frac{\pi}{180}\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}$
$\displaystyle =\frac{\pi}{180}$
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$\displaystyle \textbf{Question 10. }\text{Write the value of }\lim_{x\to0^-}\frac{\sin x}{\sqrt{x}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{x}\text{ is not defined for }x<0\text{ in real numbers.}$
$\displaystyle \therefore \lim_{x\to0^-}\frac{\sin x}{\sqrt{x}}\text{ does not exist in }R$
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$\displaystyle \textbf{Question 11. }\text{Write the value of }\lim_{x\to0}\frac{\sin x}{\sqrt{1+x}-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{\sin x}{\sqrt{1+x}-1}\cdot\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$
$\displaystyle =\frac{\sin x(\sqrt{1+x}+1)}{x}$
$\displaystyle =\frac{\sin x}{x}(\sqrt{1+x}+1)$
$\displaystyle \therefore \lim_{x\to0}\frac{\sin x}{\sqrt{1+x}-1}=1(1+1)=2$
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$\displaystyle \textbf{Question 12. }\text{Write the value of }\lim_{x\to-\infty}\left(3x+\sqrt{9x^2-x}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \lim_{x\to-\infty}\left(3x+\sqrt{9x^2-x}\right)$
$\displaystyle =\lim_{x\to-\infty}\frac{(3x+\sqrt{9x^2-x})(\sqrt{9x^2-x}-3x)}{\sqrt{9x^2-x}-3x}$
$\displaystyle =\lim_{x\to-\infty}\frac{9x^2-x-9x^2}{\sqrt{9x^2-x}-3x}$
$\displaystyle =\lim_{x\to-\infty}\frac{-x}{\sqrt{9x^2-x}-3x}$
$\displaystyle =\lim_{x\to-\infty}\frac{-x}{-x\sqrt{9-\frac{1}{x}}-3x}$
$\displaystyle =\lim_{x\to-\infty}\frac{-x}{-x\left(\sqrt{9-\frac{1}{x}}+3\right)}$
$\displaystyle =\lim_{x\to-\infty}\frac{1}{\sqrt{9-\frac{1}{x}}+3}$
$\displaystyle =\frac{1}{6}$
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$\displaystyle \textbf{Question 13. }\text{Write the value of }\lim_{n\to\infty}\frac{n!+(n+1)!}{(n+1)!+(n+2)!}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{n!+(n+1)!}{(n+1)!+(n+2)!}$
$\displaystyle =\frac{n!\{1+(n+1)\}}{(n+1)!\{1+(n+2)\}}$
$\displaystyle =\frac{n!(n+2)}{(n+1)!(n+3)}$
$\displaystyle =\frac{n+2}{(n+1)(n+3)}$
$\displaystyle \therefore \lim_{n\to\infty}\frac{n!+(n+1)!}{(n+1)!+(n+2)!}=0$
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$\displaystyle \textbf{Question 14. }\text{Write the value of }\lim_{x\to\pi/2}\frac{2x-\pi}{\cos x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x-\frac{\pi}{2}=t$
$\displaystyle \therefore 2x-\pi=2t$
$\displaystyle \cos x=\cos\left(\frac{\pi}{2}+t\right)=-\sin t$
$\displaystyle \therefore \lim_{x\to\pi/2}\frac{2x-\pi}{\cos x}=\lim_{t\to0}\frac{2t}{-\sin t}$
$\displaystyle =-2$
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$\displaystyle \textbf{Question 15. }\text{Write the value of }\lim_{n\to\infty}\frac{1+2+3+\cdots+n}{n^2}.$
$\displaystyle \text{Answer:}$
$\displaystyle 1+2+3+\cdots+n=\frac{n(n+1)}{2}$
$\displaystyle \therefore \frac{1+2+3+\cdots+n}{n^2}=\frac{n(n+1)}{2n^2}$
$\displaystyle =\frac{n+1}{2n}$
$\displaystyle =\frac{1}{2}\left(1+\frac{1}{n}\right)$
$\displaystyle \therefore \lim_{n\to\infty}\frac{1+2+3+\cdots+n}{n^2}=\frac{1}{2}$
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