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Maths MCQ

Limits · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 245 questions.

Questions · Page 2 of 5

MCQ

MCQ 512 Marks
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)=$
  • A
    $-\frac{1}{2}$
  • ✓
    $\frac{1}{2}$
  • C
    1
  • D
    -1
Answer
Correct option: B.
$\frac{1}{2}$
(B)
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)$
$=\lim _{x \rightarrow \infty} \frac{x^2+x+1-x^2-1}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}$
$=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x^2}}}$
$=\frac{1}{1+1}=\frac{1}{2}$
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MCQ 522 Marks
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+1}-x\right)$ is equal to
  • A
    1
  • B
    -1
  • ✓
    $0$
  • D
    2
Answer
Correct option: C.
$0$
(C)
On rationalising, we get
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+1}-x\right)=\lim _{x \rightarrow \infty} \frac{x^2+1-x^2}{\sqrt{x^2+1}+x}$
$=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{x^2+1}+x}=0$
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MCQ 532 Marks
$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}=$
  • A
    $\frac{3}{2}$
  • ✓
    $2 \sqrt{2}$
  • C
    2
  • D
    $\sqrt{2}$
Answer
Correct option: B.
$2 \sqrt{2}$
(B)
$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}$
$=\lim _{x \rightarrow \infty}\left(\frac{2+\frac{3}{x}+\frac{4}{x^2}}{1-\frac{3}{x}+\frac{5}{x^2}}\right)^{\left(\frac{3+\frac{1}{|x|}}{2-\frac{1}{|x|}}\right)}$
$=(2)^{\frac{2}{2}}=2 \sqrt{2}$
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MCQ 542 Marks
$\lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^5}{(2 x+3)^{45}}=$
  • A
    16
  • B
    24
  • ✓
    32
  • D
    8
Answer
Correct option: C.
32
(C)
$\lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^5}{(2 x+3)^{45}}=\lim _{x \rightarrow \infty} \frac{\left(2+\frac{1}{x}\right)^{40}\left(4-\frac{1}{x}\right)^5}{\left(2+\frac{3}{x}\right)^{45}}$
$=\frac{2^{40} \cdot 4^5}{2^{45}}=2^5=32$
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MCQ 552 Marks
The value of $\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x+\log _0(1-x)}{x^3}$ is
  • A
    -1
  • B
    $\frac{1}{2}$
  • ✓
    $-\frac{1}{2}$
  • D
    1
Answer
Correct option: C.
$-\frac{1}{2}$
(C)
$\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x+\log _e(1-x)}{x^3}$
$=\lim _{x \rightarrow 0} \frac{1}{x^3}\left[1+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots.\right)\right.$ $\left.-\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots.\right)-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots.\right)\right]$
$=\lim _{x \rightarrow 0} \frac{-x^3\left(\frac{1}{3!}+\frac{1}{3}\right)-x^4\left(\frac{1}{4!}+\frac{1}{4}\right)+\ldots}{x^3}$
$=\lim _{x \rightarrow 0}\left[-\left(\frac{1}{3!}+\frac{1}{3}\right)-x\left(\frac{1}{4!}+\frac{1}{4}\right)+\ldots.\right]$
$=-\frac{1}{2}$
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MCQ 562 Marks
The value of $\lim _{x \rightarrow 0} \frac{ e ^x+\log (1+x)-(1-x)^{-2}}{x^2}$ is equal to
  • A
    $0$
  • ✓
    -3
  • C
    -1
  • D
    infinity
Answer
Correct option: B.
-3
(B)
$\lim _{x \rightarrow 0} \frac{ e ^x+\log (1+x)-(1-x)^{-2}}{x^2}$
$=\lim _{x \rightarrow 0} \frac{1}{x^2}\left[\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\right)\right.$ $\left.\left.+\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots.\right)\right]-\left(1+2 x+3 x^2+4 x^3+\ldots.\right)\right]$
$=\lim _{x \rightarrow 0}\left(\frac{-3 x^2+x^3\left(\frac{1}{3!}+\frac{1}{3}-4\right)+\ldots .}{x^2}\right)$
$=\lim _{x \rightarrow 0}\left(-3+x\left(\frac{1}{3!}+\frac{1}{3}-4\right)+\ldots\right)=-3$
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MCQ 572 Marks
The value of $\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e +\frac{1}{2} ex }{x^2}$ is
  • ✓
    $\frac{11 e }{24}$
  • B
    $\frac{-11 e}{24}$
  • C
    $\frac{ e }{24}$
  • D
    none of these
Answer
Correct option: A.
$\frac{11 e }{24}$
(A)
$(1+x)^{\frac{1}{x}}= e ^{\frac{1}{x}[\log (1+x)]}= e ^{\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\right)}$
$= e ^{1-\frac{x}{2}+\frac{x^2}{3}-\ldots}= e . e ^{-\frac{x}{2}+\frac{x^2}{3}-\ldots}$
$=e\left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)+\frac{1}{2!}\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)^2+\ldots.\right]$
$= e \left[1-\frac{x}{2}+\frac{11}{24} x^2-\ldots.\right]$
$\therefore \quad \lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e +\frac{ e x}{2}}{x^2}=\frac{11 e }{24}$
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MCQ 582 Marks
$\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e }{x}$ equals
  • A
    $\frac{\pi}{2}$
  • B
    $0$
  • C
    $\frac{2}{ e }$
  • ✓
    $-\frac{ e }{2}$
Answer
Correct option: D.
$-\frac{ e }{2}$
(D)
$(1+x)^{\frac{1}{x}}= e ^{\frac{1}{x}[\log (1+x)]}$
$= e ^{\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right)}= e ^{\left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots\right)}$
$=e \cdot e^{\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots\right)}$
$= e \left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots.\right)\right.$ $\left.+\frac{\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}\right)}{2!}+\ldots.\right]$
$= e -\frac{ ex }{2}+\frac{1 le }{24} x^2-\ldots$
$\therefore \quad \lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e }{x}=\lim _{x \rightarrow 0}\left(\frac{-\frac{ e x}{2}+\frac{11 e }{24} x^2-\ldots .}{x}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{- e }{2}+\frac{11 e }{24} x-\ldots.\right)=-\frac{ e }{2}$
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MCQ 592 Marks
The value of $\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}$ is
  • A
    $e^{-\frac{1}{3}}$
  • B
    $e^{\frac{1}{5}}$
  • ✓
    $e^{-\frac{1}{6}}$
  • D
    $e ^{\frac{1}{2}}$
Answer
Correct option: C.
$e^{-\frac{1}{6}}$
(C)
$\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}=\lim _{x \rightarrow 0}\left(\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+. .}{x}\right)^{\frac{1}{x^2}}$
$=\lim _{x \rightarrow 0}\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\ldots\right)^{\frac{1}{x^2}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$=e^{\lim _{x \rightarrow 0} \frac{1}{x^2}\left(\frac{-x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\ldots\right)}$
$= e ^{-\frac{1}{3!}}= e ^{-\frac{1}{6}}$
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MCQ 602 Marks
The value of $\lim _{x \rightarrow 0}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}^{\frac{1}{x}}$ is
  • A
    e
  • ✓
    $e ^2$
  • C
    $\frac{2}{ e }$
  • D
    $\frac{1}{ e ^2}$
Answer
Correct option: B.
$e ^2$
(B)
$\lim _{x \rightarrow 0}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(1+\frac{2 \tan x}{1-\tan x}\right)^{\frac{1}{x}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$ =e^{\lim _{x \rightarrow 0} \frac{2 \tan x}{1-\tan x} \cdot \frac{1}{x}}$
$= e ^{\lim _{x \rightarrow 0} \frac{2}{1-\tan x} \times \frac{\tan x}{x}}$
$= e ^2$
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MCQ 612 Marks
Let $p =\lim _{x \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{2 x}}$, then $\log p$ is equal to
  • A
    1
  • ✓
    $\frac{1}{2}$
  • C
    $\frac{1}{4}$
  • D
    2
Answer
Correct option: B.
$\frac{1}{2}$
(B)
$\lim _{x \rightarrow 0}\left\{\frac{ a ^x+ b ^x+ c ^x}{3}\right\}^{\frac{1}{x}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$=e^{\lim _{x \rightarrow 0} \frac{1}{2 x} \tan ^2 \sqrt{x}}$
$= e ^{\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)^2}$
$\therefore p = e ^{\frac{1}{2}}$
$\Rightarrow \log p =\frac{1}{2}$
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MCQ 622 Marks
The value of $\lim _{x \rightarrow 0}\left\{\frac{a^x+b^x+c^x}{3}\right\}^{1 / x}$ is
  • A
    abc
  • ✓
    $(a b c)^{\frac{1}{3}}$
  • C
    $\frac{1}{3} a b c$
  • D
    $\frac{1}{3}$
Answer
Correct option: B.
$(a b c)^{\frac{1}{3}}$
(B)
$\lim _{x \rightarrow 0}\left\{\frac{ a ^x+ b ^x+ c ^x}{3}\right\}^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left\{1+\frac{ a ^x+ b ^x+ c ^x-3}{3}\right\}^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left\{1+\frac{\left( a ^x-1\right)+\left( b ^x-1\right)+\left( c ^x-1\right)}{3}\right\}^{\frac{1}{x}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$= e ^{\lim _{x \rightarrow 0}\left(\frac{ a ^x-1}{3 x}+\frac{ b ^x-1}{3 x}+\frac{ c ^x-1}{3 x}\right)} $
$= e ^{\log ( abc )^{\frac{1}{3}}}=( abc )^{\frac{1}{3}}$
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MCQ 632 Marks
The value of $\lim _{x \rightarrow 1}\left(\log _3 3 x\right)^{\operatorname{log}_x 3}$ is
  • ✓
    e
  • B
    $\frac{1}{ e }$
  • C
    1
  • D
    $-\frac{1}{e}$
Answer
Correct option: A.
e
(A)
$\lim _{x \rightarrow 1}\left(\log _3 3 x\right)^{\log _x 3}=\lim _{x \rightarrow 1}\left(\log _3 3+\log _3 x\right)^{\log _x 3}$
$=\lim _{x \rightarrow 1}\left(1+\log _3 x\right)^{\frac{1}{\log _3 x}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$= e ^{\lim _{x \rightarrow 1} \log _3 x \times \frac{1}{\log _3 x}}$
$= e ^1= e$
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MCQ 642 Marks
$\lim _{x \rightarrow e}(\log x)^{\frac{1}{1-\log x}}=$
  • A
    $0$
  • B
    e
  • ✓
    $e ^{-1}$
  • D
    $e ^2$
Answer
Correct option: C.
$e ^{-1}$
(C)
$\lim _{x \rightarrow e }(\log x)^{\frac{1}{1-\log x}}$
$=\lim _{x \rightarrow e }\left\{1+\log _{ e } x-1\right\}^{\frac{1}{1-\log x}}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$=e^{\lim _{x \rightarrow e}\left(\log _{e} x-1\right) \times \frac{1}{1-\log _{e} x}}$
$= e ^{-1}$
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MCQ 652 Marks
$\lim _{x \rightarrow 0^{+}}\left(e^x+x\right)^{1 / x}$
  • A
    does not exist finitely
  • B
    is 1
  • ✓
    is $e ^2$
  • D
    is 2
Answer
Correct option: C.
is $e ^2$
(C)
$\lim _{x \rightarrow 0^{+}}\left( e ^x+x\right)^{1 / x}$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$=e^{\lim _{x \rightarrow 0^{+}}\left(\frac{e^x-1+x}{x}\right)}$
$= e ^{\lim _{x \rightarrow 0^{+}}\left(1+\frac{ e ^x-1}{x}\right)}$
$= e ^{1+1}$
$= e ^2$
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MCQ 662 Marks
If $\lim _{x \rightarrow 0}(1+ a x)^{\frac{b}{x}}= e ^2$, where a and b are natural numbers, then
  • A
    a=4, b=2
  • B
    a=8, b=4
  • C
    a=16, b=8
  • ✓
    none of these
Answer
Correct option: D.
none of these
(D)
$\lim _{x \rightarrow 0}(1+a x)^{\frac{b}{x}}=e^2$
If $\lim _{x \rightarrow a } f (x)=1$ and $\lim _{x \rightarrow a } g (x)=\infty$, then
$\lim _{x \rightarrow a }[ f (x)]^{ g (x)}= e ^{\lim _{x \rightarrow a } g (x)[ f (x)-1]}$
$\Rightarrow e ^{\lim _{x \rightarrow 0} \frac{ a x \times b }{x}}= e ^2$
$\therefore e ^{ ab }= e ^2$
$\Rightarrow a b=2$
None of the options satisfy this equation.
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MCQ 672 Marks
$\lim _{x \rightarrow 0} x^x=$
  • A
    $0$
  • ✓
    1
  • C
    e
  • D
    None of these
Answer
Correct option: B.
1
(B)
$\lim _{x \rightarrow 0} x^r=\lim _{x \rightarrow 0} 0^{\log _e x^x}$
$= e ^{\lim _{x \rightarrow 0} x \log x}$
$= e ^{\lim _{x \rightarrow 0} \frac{\log x}{\frac{1}{x}}}$
$= e ^0$
$=1$
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MCQ 682 Marks
If $\lim _{x \rightarrow 0} \frac{\operatorname{are}^x-b \log (1+x)}{x^2}=3$, then the values of $a$, b are respectively
  • ✓
    2,2
  • B
    1,2
  • C
    2,1
  • D
    2,0
Answer
Correct option: A.
2,2
(A)
$\lim _{x \rightarrow 0} \frac{\operatorname{axe}^x- b \log (1+x)}{x^2}=3$
Applying L-Hospital's rule on L.H.S., we get
$\lim _{x \rightarrow 0}\left[\frac{ae^x+\operatorname{axe}^x-\frac{b}{1+x}}{2 x}\right]=3$ ...(i)
$\Rightarrow a +0- b =0$
$\Rightarrow a=b$ ...(ii)
From (i),
$\lim _{x \rightarrow 0} \frac{ a \left( e ^x+x e ^x-\frac{1}{1+x}\right)}{2 x}=3$
Applying L-Hospital's rule on L.H.S., we get
$a \lim _{x \rightarrow 0} \frac{ e ^x+ e ^x+x e ^x+\frac{1}{(1+x)^2}}{2}=3$
$\Rightarrow \frac{3 a}{2}=3 \Rightarrow a=2$
From (ii), b = 2
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MCQ 692 Marks
$\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2}$ equals
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{2}$
  • ✓
    $\frac{3}{2}$
Answer
Correct option: D.
$\frac{3}{2}$
(D)
Let $y=\lim _{x \rightarrow 0} \frac{x e ^x-\log (1+x)}{x^2}$
Applying L-Hospital's rule, we get
$y=\lim _{x \rightarrow 0} \frac{ e ^x+x e ^x-\frac{1}{1+x}}{2 x}$
$=\lim _{x \rightarrow 0} \frac{1}{2}\left[ e ^x+ e ^x+x e ^x+\frac{1}{(1+x)^2}\right]$
$=\frac{1}{2}(1+1+0+1)=\frac{3}{2}$
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MCQ 702 Marks
$\lim _{x \rightarrow 0} \frac{\sin x+\log (1-x)}{x^2}$ is equal to
  • A
    $0$
  • B
    $\frac{1}{2}$
  • ✓
    $-\frac{1}{2}$
  • D
    None of these
Answer
Correct option: C.
$-\frac{1}{2}$
(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\cos x-\frac{1}{1-x}}{2 x}=\lim _{x \rightarrow 0} \frac{-\sin x-\frac{1}{(1-x)^2}}{2}=-\frac{1}{2}$
Alternate method:
$\lim _{x \rightarrow 0} \frac{\sin x+\log (1-x)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right)}{x^2}$ $+\lim _{x \rightarrow 0} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots\right)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\frac{-x^2}{2}-x^3\left(\frac{1}{3!}+\frac{1}{3}\right)-\frac{x^4}{4} \ldots}{x^2}=-\frac{1}{2}$
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MCQ 712 Marks
$\lim _{x \rightarrow 0} \frac{\log (\cos x)}{x^2}$ is equal to
  • A
    $0$
  • B
    1
  • C
    $\frac{1}{2}$
  • ✓
    $-\frac{1}{2}$
Answer
Correct option: D.
$-\frac{1}{2}$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\log (\cos x)}{x^2}=\lim _{x \rightarrow 0} \frac{-\tan x}{2 x}$
$=\lim _{x \rightarrow 0} \frac{-\sec ^2 x}{2}=\frac{-1}{2}$
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MCQ 722 Marks
The value of $\lim _{x \rightarrow 1} \frac{\log x}{\sin \pi x}$ is
  • A
    $\frac{1}{\pi}$
  • B
    $-\pi$
  • C
    $\pi$
  • ✓
    $-\frac{1}{\pi}$
Answer
Correct option: D.
$-\frac{1}{\pi}$
(D)
$\lim _{x \rightarrow 1} \frac{\log x}{\sin \pi x}=\lim _{x \rightarrow 1} \frac{\log [1+(x-1)]}{\sin (\pi-\pi x)}$
$=\lim _{x \rightarrow 1} \frac{\log [1+(x-1)]}{\sin \pi(1-x)}$
$=\lim _{x \rightarrow 1} \frac{\log [1+(x-1)]}{x-1} \times \frac{x-1}{\sin \pi(1-x)}$
$=-\frac{1}{\pi} \lim _{x \rightarrow 1} \frac{\log [1+(x-1)]}{x-1} \times \frac{\pi(1-x)}{\sin \pi(1-x)}=-\frac{1}{\pi}$
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MCQ 732 Marks
The value of $\lim _{x \rightarrow 0} \frac{\left(4^{\prime}-1\right)^3}{\sin \frac{x^2}{4} \log (1+3 x)}$ is
  • A
    $\frac{4}{3}\left(\log _e 4\right)^2$
  • ✓
    $\frac{4}{3}\left(\log _e 4\right)^3$
  • C
    $\frac{3}{2}\left(\log _{e} 4\right)^2$
  • D
    $\frac{3}{2}\left(\log _e 4\right)^4$
Answer
Correct option: B.
$\frac{4}{3}\left(\log _e 4\right)^3$
(B)
$\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x^2}{4} \log (1+3 x)}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{4^x-1}{x}\right)^3}{\frac{1}{4}\left\{\frac{\sin \left(x^2 / 4\right)}{\left(x^2 / 4\right)}\right\} \times \frac{\log (1+3 x)}{3 x} \times 3}=\frac{4}{3}\left(\log _e 4\right)^3$
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MCQ 742 Marks
If $\lim _{x \rightarrow 0} \frac{\log (x+a)-\log a}{x}+k \lim _{x \rightarrow 0} \frac{\log x-1}{x-c}=1$, then the value of $k$ is
  • A
    $1-\frac{1}{a}$
  • B
    $e(1-a)$
  • ✓
    $e\left(1-\frac{1}{a}\right)$
  • D
    $e(1+a)$
Answer
Correct option: C.
$e\left(1-\frac{1}{a}\right)$
(C)
$\lim _{x \rightarrow 0} \frac{\log (x+ a )-\log a }{x}+ k \lim _{x \rightarrow c } \frac{\log x-1}{x- e }=1$
Applying L-Hospital's rule on L.H.S., we get
$\frac{1}{a}+\frac{k}{e}=1 \Rightarrow k=e\left(1-\frac{1}{a}\right)$
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MCQ 752 Marks
If $\lim _{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=k$, then the value of, k is
  • A
    $0$
  • B
    $-\frac{1}{3}$
  • ✓
    $\frac{2}{3}$
  • D
    $-\frac{2}{3}$
Answer
Correct option: C.
$\frac{2}{3}$
(C)
$\lim _{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}= k$
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\frac{1}{3+x}+\frac{1}{3-x}}{1}= k$
$\therefore \quad k =\frac{2}{3}$
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MCQ 762 Marks
$\lim _{x \rightarrow 0} \frac{2 \log (1+x)-\log (1+2 x)}{x^2}$ is equal to
  • A
    $0$
  • ✓
    1
  • C
    -1
  • D
    2
Answer
Correct option: B.
1
(B)
$\lim _{x \rightarrow 0} \frac{2 \log (1+x)-\log (1+2 x)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\log \left\{\frac{(1+x)^2}{1+2 x}\right\}}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{x^2}{1+2 x}\right)}{x^2} \times \frac{1}{1+2 x}=1$
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MCQ 772 Marks
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=$
  • A
    1
  • B
    -1
  • C
    $0$
  • ✓
    $-\frac{1}{2}$
Answer
Correct option: D.
$-\frac{1}{2}$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)}$
Again applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2}$
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MCQ 782 Marks
$\lim _{x \rightarrow 0}\left[\frac{1}{x}-\frac{\log (1+x)}{x^2}\right]=$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{-1}{2}$
  • C
    1
  • D
    -1
Answer
Correct option: A.
$\frac{1}{2}$
(A)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0}\left[\frac{x-\log (1+x)}{x^2}\right]=\lim _{x \rightarrow 0} \frac{1-\frac{1}{1+x}}{2 x}$
$=\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{1}{1+x}\right)^2=\frac{1}{2}$
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MCQ 792 Marks
$\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=$
  • ✓
    $e^3$
  • B
    $e^6$
  • C
    $e^9$
  • D
    $e ^{-3}$
Answer
Correct option: A.
$e^3$
(A)
$\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(\frac{1+\frac{5 x}{3}}{1-\frac{4 x}{3}}\right)^{\frac{1}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{5 x}{3}\right)^{\frac{3}{5 x}}\right]^{\frac{5}{3}}}{\lim _{x \rightarrow 0}\left[\left(1-\frac{4 x}{3}\right)^{\frac{-3}{4 x}}\right]^{\frac{-4}{3}}}$
$=\frac{ e ^{\frac{5}{3}}}{ e ^{\frac{-4}{3}}}= e ^3$
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MCQ 802 Marks
$\lim _{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{1 / x^2}=$
  • ✓
    $e^2$
  • B
    e
  • C
    $e ^{-2}$
  • D
    $e ^{-1}$
Answer
Correct option: A.
$e^2$
(A)
$\lim _{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{1 / x^2}=\frac{\lim _{x \rightarrow 0}\left[\left(1+5 x^2\right)^{1 / 5 x^2}\right]^5}{\lim _{x \rightarrow 0}\left[\left(1+3 x^2\right)^{1 / 3 x^2}\right]^7}$
$=\frac{ e ^5}{ e ^3} \quad \ldots .\left[\because \lim _{x \rightarrow 0}(1+x)^{1 / x}= e \right]$
$= e ^2$
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MCQ 812 Marks
If $\lim _{x \rightarrow a} \frac{ a ^x-x^a}{x^x- a ^x}=-1$, then
  • ✓
    $a=1$
  • B
    $a =0$
  • C
    $a = e$
  • D
    none of these
Answer
Correct option: A.
$a=1$
(A)
$\lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$
Applying L-Hospital's rule on L.H.S., we get
$\lim _{x \rightarrow a} \frac{a^x \log _e a-a x^{a-1}}{x^x\left(1+\log _e x\right)}=-1$
$\Rightarrow \frac{a^a \log _e a-a \cdot a^{a-1}}{a^a\left(1+\log _e a\right)}=-1$
$\Rightarrow \frac{\log _{ e } a -1}{\log _{ e } a +1}=-1$
$\Rightarrow \log _{ e } a -1=-\log _{ e } a -1$
$\Rightarrow 2 \log _{ e } a =0 \Rightarrow a = e ^0=1$
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MCQ 822 Marks
$\lim _{x \rightarrow 0} \frac{ e ^{1 / x}-1}{ e ^{1 / x}+1}=$
  • A
    $0$
  • B
    1
  • C
    -1
  • ✓
    Does not exist
Answer
Correct option: D.
Does not exist
(D)
Let $f(x)=\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$, then
$\lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{h \rightarrow 0}\left(\frac{e^{1 / h}-1}{e^{1 / h}+1}\right)=\lim _{h \rightarrow 0} \frac{e^{1 / h}\left(1-\frac{1}{e^{1 / h}}\right)}{e^{1 / h}\left(1+\frac{1}{e^{1 / h}}\right)}=1$
Similarly, $\lim _{x \rightarrow 0^{-}} f (x)=-1$.
Hence, limit does not exist.
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MCQ 832 Marks
$\lim _{x \rightarrow 0^{+}} \frac{x e ^{1 / x}}{1+ e ^{1 / x}}=$
  • ✓
    $0$
  • B
    1
  • C
    $\infty$
  • D
    None of these
Answer
Correct option: A.
$0$
(A)
$\lim _{x \rightarrow 0^{+}} \frac{x}{1+ e ^{-1 / x}}-0$ as $e ^{-1 / x} \rightarrow 0$ when $x \rightarrow 0^{+}$
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MCQ 842 Marks
If $f(x)=\frac{\sin \left(e^{x-3}-1\right)}{\log (x-2)}$, then $\lim _{x \rightarrow 3} f(x)=$
  • A
    -2
  • B
    -1
  • ✓
    1
  • D
    $0$
Answer
Correct option: C.
1
(C)
$\lim _{x \rightarrow 3} f (x)=\lim _{x \rightarrow 3} \frac{\sin \left( e ^{x-3}-1\right)}{\log (x-2)}$
$=\lim _{h \rightarrow 0} \frac{\sin \left(e^h-1\right)}{\log (1+h)}$
$=\lim _{h \rightarrow 0} \frac{\sin \left(e^h-1\right)}{e^h-1} \cdot \frac{e^h-1}{h} \cdot \frac{h}{\log (1+h)}=1$
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MCQ 852 Marks
The value of $\lim _{x \rightarrow 2} \frac{ e ^{3 x-6}-1}{\sin (2-x)}$ is
  • A
    $\frac{3}{2}$
  • B
    3
  • ✓
    -3
  • D
    -1
Answer
Correct option: C.
-3
(C)
$\lim _{x \rightarrow 2} \frac{ e ^{3 x-6}-1}{\sin (2-x)}=\lim _{x \rightarrow 2} \frac{ e ^{3(x-2)}-1}{3(x-2)} \times \frac{1}{\frac{\sin (2-x)}{-3(2-x)}}$
$=1 \times-3$
$\ldots\left[\because \lim _{x \rightarrow 0} \frac{ e ^x-1}{x}=1, \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=-3$
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MCQ 862 Marks
$\lim _{x \rightarrow 0} \frac{x\left( e ^{\sin x}-1\right)}{1-\cos x}$ is equal to
  • A
    1
  • B
    $0$
  • ✓
    2
  • D
    $\frac{1}{2}$
Answer
Correct option: C.
2
(C)
$\lim _{x \rightarrow 0} \frac{x\left( e ^{\sin x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \frac{\frac{ e ^{\sin x}-1}{x}}{\frac{1-\cos x}{x^2}}$
$=\left(\lim _{x \rightarrow 0} \frac{ e ^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x}\right) \times 2=2$
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MCQ 872 Marks
$\lim _{x \rightarrow 0} \frac{ e ^{\tan x}- e ^2}{\tan x-x}=$
  • ✓
    1
  • B
    e
  • C
    $e^{-1}$
  • D
    $0$
Answer
Correct option: A.
1
(A)
$\lim _{x \rightarrow 0} \frac{ e ^{\tan x}- e ^x}{\tan x-x}=\lim _{x \rightarrow 0} \frac{ e ^x\left( e ^{\tan x-x}-1\right)}{\tan x-x}$
$=\lim _{x \rightarrow 0} e ^x\left(\frac{ e ^{\tan x-x}-1}{\tan x-x}\right)$
$= e ^0 \times 1$
$=1$
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MCQ 882 Marks
The value of $\lim _{x \rightarrow 0} \frac{ e ^x- e ^{\sin x}}{2(x-\sin x)}$ is
  • A
    $-\frac{1}{2}$
  • ✓
    $\frac{1}{2}$
  • C
    1
  • D
    $\frac{3}{2}$
Answer
Correct option: B.
$\frac{1}{2}$
(B)
$\lim _{x \rightarrow 0} \frac{ e ^x- e ^{\sin x}}{2(x-\sin x)}=\frac{1}{2} \lim _{x \rightarrow 0} e ^{\sin x}\left(\frac{ e ^{x-\sin x}-1}{(x-\sin x)}\right)$
$=\frac{1}{2} \times e ^0 \times 1=\frac{1}{2}$
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MCQ 892 Marks
$\lim _{\theta \rightarrow 0} \frac{e^\theta+e^{-\theta}-2}{\sin ^2 \theta}$ is equal to
  • ✓
    1
  • B
    $0$
  • C
    -1
  • D
    $\sqrt{-2}$
Answer
Correct option: A.
1
(A)
$\lim _{\theta \rightarrow 0} \frac{e^\theta+e^{-\theta}-2}{\sin ^2 \theta}=\lim _{\theta \rightarrow 0} \frac{e^\theta+\frac{1}{e^\theta}-2}{\sin ^2 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{\left[e^\theta-1\right]^2}{e^\theta \sin ^2 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{\frac{\left(e^\theta-1\right)^2}{\theta^2}}{e^\theta \cdot \frac{\sin ^2 \theta}{\theta^2}}$
$=\frac{[\log e ]^2}{1}-1$
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MCQ 902 Marks
$\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}-2}{x^2}=$
  • ✓
    1
  • B
    -1
  • C
    $\frac{1}{2}$
  • D
    $-\frac{1}{2}$
Answer
Correct option: A.
1
(A)
$\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}-2}{x^2}=\lim _{x \rightarrow 0} \frac{ e ^x+\frac{1}{ e ^x}-2}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\left( e ^x\right)^2-2 e ^x+1}{x^2 e ^x}=\lim _{x \rightarrow 0}\left(\frac{ e ^x-1}{x}\right)^2 \cdot \frac{1}{ e ^x}$
$=(1)^2 \cdot \frac{1}{ e ^0}=1$
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MCQ 912 Marks
The value of $\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$ is
  • A
    $\sqrt{5}(\log 3)^2$
  • B
    $8 \sqrt{5} \log 3$
  • C
    $16 \sqrt{5} \log 3$
  • ✓
    $8 \sqrt{5}(\log 3)^2$
Answer
Correct option: D.
$8 \sqrt{5}(\log 3)^2$
(D)
$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$
$=\lim _{x \rightarrow 0} \frac{\left(9^x \cdot 3^x-9^x-3^x+1\right)}{\sqrt{5}-\sqrt{4+\cos x}} \times \frac{\sqrt{5}+\sqrt{4+\cos x}}{\sqrt{5}+\sqrt{4+\cos x}}$
$=\lim _{x \rightarrow 0} \frac{\left[9^x\left(3^x-1\right)-1\left(3^x-1\right)\right][\sqrt{5}+\sqrt{4+\cos x}]}{1-\cos x}$
$=\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(9^x-1\right)(\sqrt{5}+\sqrt{4+\cos x})}{2 \sin ^2 \frac{x}{2}}$
$=\lim _{x \rightarrow 0} \frac{\frac{3^x-1}{x} \cdot \frac{9^x-1}{x}(\sqrt{5}+\sqrt{4+\cos x})}{2 \cdot \frac{\sin ^2 \frac{x}{2}}{4 \times \frac{x^2}{4}}}$
$=2 \cdot \log 3 \cdot \log 9 \cdot(2 \sqrt{5})=8 \sqrt{5}(\log 3)^2$
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MCQ 922 Marks
$\lim _{x \rightarrow \pi / 2} \frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}=$
  • ✓
    $\log a$
  • B
    $\log 2$
  • C
    a
  • D
    $\log x$
Answer
Correct option: A.
$\log a$
(A)
$\lim _{x \rightarrow \pi / 2}\left(\frac{ a ^{\cot x}- a ^{\cos x}}{\cot x-\cos x}\right)=\lim _{x \rightarrow \pi / 2} a ^{\cos x}\left(\frac{ a ^{\cot x-\cos x}-1}{\cot x-\cos x}\right)$
$= a ^{\cos \frac{\pi}{2}} \lim _{x>\pi / 2}\left(\frac{ a ^{\cot x-\cos x}-1}{\cot x-\cos x}\right)$
$=1 \cdot \log a =\log a$
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MCQ 932 Marks
$\lim _{x \rightarrow 0}\left(\frac{2^x-2^x \cos x+\cos x-1}{x^3}\right)$ is equal to
  • A
    $2 \log 2$
  • ✓
    $\frac{\log 2}{2}$
  • C
    $0$
  • D
    $\log 2$
Answer
Correct option: B.
$\frac{\log 2}{2}$
(B)
$\lim _{x \rightarrow 0} \frac{2^x-2^x \cos x+\cos x-1}{x^3}$
$=\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)\left(\frac{1-\cos x}{x^2}\right)$
$=(\log 2) \frac{1}{2}=\frac{\log 2}{2}$
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MCQ 942 Marks
$\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ is equal to
  • A
    $\log 2$
  • B
    $\frac{1}{2} \log 2$
  • ✓
    $2 \log 2$
  • D
    $\frac{1}{2}$
Answer
Correct option: C.
$2 \log 2$
(C)
$\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x^2}{1-\cos x} \times \lim _{x \rightarrow 0} \frac{2^x-1}{x}$
$=2 \log 2$
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MCQ 952 Marks
$\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=$
  • A
    $\log 2$
  • ✓
    $\log 4$
  • C
    $\log \sqrt{2}$
  • D
    $3 \log 2$
Answer
Correct option: B.
$\log 4$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=\lim _{x \rightarrow 0} \frac{2^x \log 2}{\frac{1}{2}(1+x)^{-1 / 2}}$
$=2 \log 2=\log 4$
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MCQ 962 Marks
$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}=$
  • ✓
    $\log \left(\frac{2}{3}\right)$
  • B
    $\frac{1}{2} \log \left(\frac{3}{2}\right)$
  • C
    $\frac{1}{2} \log \left(\frac{2}{3}\right)$
  • D
    $\log \left(\frac{3}{2}\right)$
Answer
Correct option: A.
$\log \left(\frac{2}{3}\right)$
(A)
Applying L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x\left(4^x+9^x\right)}=\lim _{x \rightarrow 0} \frac{4^x \log 4-9^x \log 9}{\left(4^x+9^x\right)+x\left(4^x \log 4+9^x \log 9\right)}$
$=\frac{\log 4-\log 9}{2}=\frac{\log \left(\frac{2}{3}\right)^2}{2}$
$=\log \frac{2}{3}$
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MCQ 972 Marks
$\lim _{x \rightarrow 0} \frac{12^x-3^x-4^x+1}{x \sin x}$ is equal to
  • A
    $\log \sqrt{3} \cdot \log 4$
  • ✓
    $\log 4 \cdot \log 3$
  • C
    $\log 2 \cdot \log 3$
  • D
    $\log \sqrt{3} \cdot \log 2$
Answer
Correct option: B.
$\log 4 \cdot \log 3$
(B)
$\lim _{x \rightarrow 0} \frac{12^x-3^x-4^x+1}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{(4 \times 3)^x-3^x-4^x+1}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{4^x\left(3^x-1\right)-1\left(3^x-1\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)\left(3^x-1\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{4^x-1}{x}\right) \cdot\left(\frac{3^x-1}{x}\right)}{\frac{\sin x}{x}}$
$=\log 4 \cdot \log 3$
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MCQ 982 Marks
$\lim _{x \rightarrow 0} \frac{a^{3 x}-a^{2 x}-a^x+1}{x^2}$ is equal to
  • A
    $\sqrt{2} \log a^2$
  • B
    $\sqrt{2} \log a$
  • ✓
    $2(\log a)^2$
  • D
    $2(\log a)^3$
Answer
Correct option: C.
$2(\log a)^2$
(C)
$\lim _{x \rightarrow 0} \frac{ a ^{3 x}- a ^{2 x}- a ^x+1}{x^2}$
$=\lim _{x \rightarrow 0} \frac{ a ^{2 x} \cdot a ^x- a ^{2 x}- a ^x+1}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\left( a ^{2 x}-1\right)\left( a ^x-1\right)}{x^2}$
$=2 \log a \cdot \log a =2(\log a )^2$
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MCQ 992 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[1-\tan \left(\frac{x}{2}\right)\right](1-\sin x)}{\left[1+\tan \left(\frac{x}{2}\right)\right](\pi-2 x)^3}$ is
  • A
    $\frac{1}{8}$
  • B
    $0$
  • ✓
    $\frac{1}{32}$
  • D
    $\infty$
Answer
Correct option: C.
$\frac{1}{32}$
(C)
Let $l=\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right]\left[\frac{(1-\sin x)}{(\pi-2 x)^3}\right]$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right](1-\sin x)}{(\pi-2 x)^3}$
Put $\pi-2 x=\theta$
$\Rightarrow x=\frac{\pi}{2}-\frac{\theta}{2}$ and as $x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0$
$\therefore l=\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{4}\left(1-\cos \frac{\theta}{2}\right)}{\theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{4}}{\frac{\theta}{4} \times 4} \cdot \frac{2 \sin ^2 \frac{\theta}{4}}{\frac{\theta^2}{16} \times 16}$
$=\frac{1}{32} \lim _{\theta \rightarrow 0}\left[\frac{\tan \frac{\theta}{4}}{\frac{\theta}{4}} \cdot\left(\frac{\sin \frac{\theta}{4}}{\frac{\theta}{4}}\right)^2\right]$
$=\frac{1}{32}(1)(1)^2=\frac{1}{32}$
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MCQ 1002 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{24}$
  • ✓
    $\frac{1}{16}$
  • D
    $\frac{1}{8}$
Answer
Correct option: C.
$\frac{1}{16}$
(C)
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
Put $\pi-2 x=\theta$
$\Rightarrow x=\frac{\pi}{2}-\frac{\theta}{2}$ and as $x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0$
$\lim _{\theta \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-\frac{\theta}{2}\right)-\cos \left(\frac{\pi}{2}-\frac{\theta}{2}\right)}{\theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{2}-\sin \frac{\theta}{2}}{\theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}\left(1-\cos \frac{\theta}{2}\right)}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2} \cdot 2 \sin ^2 \frac{\theta}{4}}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=2 \lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2} \times 2} \cdot \frac{\sin ^2 \frac{\theta}{4}}{\left(\frac{\theta}{4}\right)^2 \times 16} \cdot \frac{1}{\cos \frac{\theta}{2}}$
$=\frac{2}{32}=\frac{1}{16}$
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MCQ - Page 2 - Maths STD 11 Science Questions - Vidyadip