If $$ f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}}, $$ then

$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 7

Evaluate

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

0%
$4$
0%
$2$
0%
$1$
0%
$\dfrac{1}{2}$

Explanation

We have,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$

We know that

$\cos 2x=1-2\sin^2x$

Therefore,

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - (1-2\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (2\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - (1-2\sin^2 (\sin^2 x))}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{{2\sin^2 (\sin^2 x)}}{{{x^4}}}$

$\mathop {\lim }\limits_{x \to 0} \cfrac{2\sin^2 (\sin^2 x)}{\sin^4 x}\times \dfrac{\sin^4 x}{x^4}$

$2\mathop {\lim }\limits_{x \to 0} \left(\cfrac{\sin^2 (\sin^2 x)}{(\sin^2 x)}\right)^2\times \dfrac{\sin^4 x}{x^4}$

We know that

$\lim_{x\to\ 0}\dfrac{\sin x}{x}=1$

Therefore,

$\Rightarrow 2\times 1\times 1$

$\Rightarrow 2$

Hence, this is the answer.

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 2}\dfrac{x-2}{\sqrt{x}-\sqrt{2}}$ .

0%
$2\sqrt{3}$
0%
$2\sqrt{2}$
0%
$2\sqrt{5}$
0%
None of these

Explanation

$\displaystyle\lim_{x\rightarrow 2}\dfrac{(x-2)}{\sqrt{x}-\sqrt{2}}$

$x\rightarrow 2$ , it is

$\dfrac{0}{0}$ Form,

So, using Factorisation

$\displaystyle\lim_{x\rightarrow 2}\dfrac{((\sqrt x)^2-(\sqrt 2)^2)}{\sqrt{x}-\sqrt{2}}=\displaystyle\lim_{x\rightarrow 2}\dfrac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})}$

$\displaystyle \lim_{x\to 2}\sqrt x +\sqrt 2=2\sqrt{2}$

$\lim _{ x\rightarrow 0 }{ \cfrac { x\left( 1+a\cos { x } \right) -b\sin { x } }{ { x }^{ 3 } } } =1$ then

0%
$a=-5/2$ , $b=-1/2$
0%
$a=-3/2$ , $b=-1/2$
0%
$a=-3/2$ , $b=-5/2$
0%
$a=-5/2$ , $b=-3/2$

$\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x }$ is :

0%
$-2$
0%
$\dfrac { -1 }{ 2 }$
0%
$\dfrac { 1 }{ 2 }$
0%
$2$

Explanation

Let

$L=\displaystyle \lim_{x\rightarrow 0} \frac { (1 - cos 2x)^2}{2x \tan x - x \tan 2x}= \lim_{x\rightarrow 0} \frac {( 2 sin x)^2}{2x \tan x- \frac {x 2 \tan x}{1 - \tan^2 x}}$

$=\displaystyle \lim_{x\rightarrow 0} \frac { { 4 sin^4 x}}{\frac{1 - \tan^2 x}{2 x \tan x - 2x \tan^3 x - 2x \tan x}}$

$=\displaystyle \lim_{x\rightarrow 0} \frac {4 sin ^4 x ( 1 - \tan^2 x)}{- 2 x \tan^3 x}$

$=\displaystyle \lim_{x\rightarrow 0} \frac {2 sin^4 x}{-x \frac {sin^3 x}{cos x}}( 1 - \tan^2 x)$

$\displaystyle =- 2 \lim_{x\rightarrow 0} \left \{ \frac {sin x . cos^3 x}{x} (1 \tan^2 x) \right \}$

$\displaystyle = -2 \left ( \lim_{x\rightarrow 0} \frac { sin x}{x} \right ) \lim_{x\rightarrow 0} \left \{ cos^3 x ( 1 - \tan^2 x) \right \}$

$L\displaystyle = -2(1)(1)$

$L\displaystyle = -2$

$\underset { x\rightarrow 0 }{ \lim } \dfrac { { 3 }^{ 2x }-{ 2 }^{ 3x } }{ x }$ is equal to

0%
$\log\dfrac { 3 }{ 2 }$
0%
$1$
0%
$\log\cfrac { 9 }{ 8 }$
0%
$0$

Explanation

We have,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}-{{2}^{3x}}}{x}$

Applying L’ Hospital rule and we get,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}\log 3\times2-{{2}^{3x}}\log 2\times(3)}{1}$

Taking limit and we get,

$={{3}^{2\times 0}}\log 3^2-{{2}^{3\times 0}}\log 2^3$

$=\log 9-\log 8$

$=\log \dfrac{9}{8}$

Hence, this is the answer.

Let

$f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2$ and

$lim_{x\rightarrow \infty} f(x)=1$ then

$f(-2)=$

0%
$1$
0%
$2$
0%
$-1$
0%
$0$

Explanation

$\underset{x\rightarrow 0}{lim} f(x)=2$

$\Rightarrow \underset{x\rightarrow 0}{lim} \dfrac{ax+b}{x+1}=2$

$\Rightarrow \dfrac{b}{1}=2$

$\therefore b=2$

Its is also given that

$\underset{x\rightarrow \infty}{lim} f(x)=1$

$\Rightarrow \underset{x\rightarrow \infty}{lim} \dfrac{ax+b}{x+1}=1$

$\Rightarrow \underset{x\rightarrow \infty}{lim}\dfrac{a+\dfrac{b}{x}}{1+\dfrac{1}{x}}=1$

$\Rightarrow \dfrac{a+0}{1+0}=1$

$\therefore a=1$

Substituting the values of a and b in

$f(x)=\dfrac{ax+b}{x+1}$ , we get,

$f(x)=\dfrac{x+2}{x+1}$

$\therefore f(-2)=\dfrac{-2+2}{-2+1}=0$

Evaluate the limit,

$\mathop {\lim }\limits_{x \to 0} \frac{{x({{(1 + x)}^{1/x}} - e)}}{{x({{(1 + {x^2})}^{1/{x^2}}} - e)}}$

0%
0
0%
1
0%
2
0%
DNE

$\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } - x + 1 } - a x - b \right) = 0,$ then the values of

$a$ and

$b$ are given by

0%
$a = - 1 , b = 1 / 2$
0%
$a = 1 , b = 1 / 2$
0%
$a = 1 , b = - 1 / 2$
0%
None of these

Explanation

We have,

$\begin{array}{l} { \lim }_{ x\to \infty }\left( { \sqrt { { x^{ 2 } }-x+1 } -ax-b } \right) =0 \\ Put\, \, x=\cfrac { 1 }{ t } \\ { \lim }_{ t\to { 0^{ + } } }\left( { \sqrt { \cfrac { 1 }{ { { t^{ 2 } } } } -\cfrac { 1 }{ t } +1 } -\cfrac { a }{ t } -6 } \right) =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { { { \left( { 1-t+{ t^{ 2 } } } \right) }^{ \cfrac { 1 }{ 2 } } }-a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-\cfrac { t }{ 2 } +\cfrac { { { t^{ 2 } } } }{ 2 } } \right) -a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-a } \right) +t\left( { -b-\cfrac { 1 }{ 2 } } \right) } }{ t } =0 \\ 1=a \\ and\, \, b=-\cfrac { 1 }{ 2 } \\ Option\, \, \, C\, \, \, is\, \, correct\, \, answer. \end{array}$

The value of

$\lim_{x\rightarrow \infty }$ y In

$(\frac{sin (x+1/y)}{sin x})$ when

$0 < x < \pi /2$ is

0%
cos x
0%
$\infty$
0%
cot x
0%
does not exist

$\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan x }{ \cos (x+\pi /4) }$ is

0%
$4$
0%
$8 \sqrt { 2 }$
0%
$8$
0%
$4 \sqrt { 2 }$

$\displaystyle\lim_{n\rightarrow\infty}\left\{\dfrac{n!}{(kn)^n}\right\}^{\dfrac{1}{n}}, k\neq 0$ , is equal to?

0%
$\dfrac{k}{e}$
0%
$\dfrac{e}{k}$
0%
$\dfrac{1}{ke}$
0%
None of these

Let

$f(x)=\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{r=0}\dfrac{x}{(rx+1)\{(r+1)x+1\}}$ , then?

0%
$f(x)$ is continuous but not differentiable at $x=0$
0%
$f(x)$ is both continuous and differentiable at $x=0$
0%
$f(x)$ is neither continuous nor differentiable at $x=0$
0%
$f(x)$ is a periodic function

$\underset { x\rightarrow 0 }{ lim } \frac { tan(sinx)-x }{ { tanx }^{ 3 } }$ is equal to

0%
$\frac { 1 }{ 6 }$
0%
$\frac { 1 }{ 3 }$
0%
$\frac { 1 }{ 2 }$
0%
1

$\displaystyle \lim _{ x\rightarrow \infty }{ \left[\dfrac{n}{n^{2}+1^{2}}+\dfrac{n}{n^{2}+2^{2}}+\dfrac{n}{n^{2}+3^{2}}+....+\dfrac{1}{n^{5}}\right] }$

0%
$\pi/4$
0%
$\tan^{-1}{(2)}$
0%
$\pi/2$
0%
$\tan^{-1}{(3)}$

The value of

$\displaystyle n\xrightarrow { lim } \infty\frac{1.n+2.(n-1)+3.(n-2)+...+n.1}{{1}^{2}+{2}^{2}+...+{n}^{2}}$ is

0%
$1$
0%
$-1$
0%
$\displaystyle \frac{1}{\sqrt{2}}$
0%
$\displaystyle \frac{1}{2}$

Explanation

Now,

$1.n+2(n-1)+3.(n-2)+....+n.1$

The general term of the above series is

${ T }_{ r }=r(n-r+1)=r(n+1-r)$

Sum of series

$=S=\sum _{ r=1 }^{ n }{ { T }_{ r } } =\sum _{ r=1 }^{ n }{ r(n+1-r) }$

$S=\sum _{ r=1 }^{ n }{ nr } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\sum _{ r=1 }^{ n }{ r } +\sum _{ r=1 }^{ n }{ r } -\sum _{ r=1 }^{ n }{ { r }^{ 2 } } \\ =n\cfrac { n(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 }$

Similarly,

${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ n }^{ 2 }=\cfrac { n(n+1)(2n+1) }{ 6 }$

$\lim _{ n\rightarrow \infty }{ \cfrac { 1.n+2(n-1)+3.(n-2)+...+n.1 }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+........{ n }^{ 2 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { { n }^{ 2 }(n+1) }{ 2 } +\cfrac { n(n+1) }{ 2 } -\cfrac { n(n+1)(2n+1) }{ 6 } }{ \cfrac { n(n+1)(2n+1) }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { n }{ 2 } +\cfrac { 1 }{ 2 } -\cfrac { 2n+1 }{ 6 } }{ \cfrac { 2n+1 }{ 6 } } } \\ \lim _{ n\rightarrow \infty }{ (\cfrac { 3n+3-2n-1 }{ 2n+1 } ) } =\lim _{ n\rightarrow \infty }{ (\cfrac { 3+\cfrac { 3 }{ n } -2-\cfrac { 1 }{ n } }{ 2+\cfrac { 1 }{ n } } ) } \\ =\cfrac { 3-2 }{ 2 } =\cfrac { 1 }{ 2 }$

$\displaystyle\lim _{ x\rightarrow 0 }{ { x }^{ 2 }{ e }^{ \sin { \frac { 1 }{ x } } } }$ equals

0%
$1$
0%
$0$
0%
$\infty$
0%
Does not exist

Evaluate:

$\underset { { x\rightarrow}\dfrac{ \pi }{ 4 } }{ lim } \dfrac { { cot }^{ 3 }x-tanx }{ cos\left( x+\pi /4 \right) } \quad$

0%
$8$
0%
$8\sqrt { 2 }$
0%
$4$
0%
$4\sqrt { 2 }$

Explanation

Using L'Hopital rule:

$\underset { { x\rightarrow}\dfrac{ \pi }{ 4 } }{ lim } \dfrac { { cot }^{ 3 }x-tanx }{ cos\left( x+\pi /4 \right) } \quad =\lim\limits_{x\to\pi/4}\dfrac{-3\cot^2x\ cosec^2\ x-\sec^2 x}{-sin(x+\pi/4)}=\dfrac{-3\times1\times(\sqrt2)^2-(\sqrt2)^2}{-1}=8$

${ lim }_{ x\rightarrow 1 }(1+cos\pi ){ cot }^{ 2 }\pi x=-----$

0%
1
0%
-1
0%
$\dfrac { 1 }{ 2 }$
0%
0

Let

$f:(0, \infty)\to R$ be a differentiable function such that

$f'(x)=2-\dfrac{f(x)}{x}$ for all

$x\in (0, \infty)$ and

$f(1)\neq 1$ . Then

0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } f'\left( \dfrac { 1 }{ x } \right) =1$
0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } xf\left( \dfrac { 1 }{ x } \right) =2$
0%
$\underset { x\rightarrow { 0 }^{ + } }{ \lim } x^{ 2 }f'\left( x \right) =0$
0%
$\left| f\left( x \right) \right| \le 2$ for all $X\in \left( 0,2 \right)$

Explanation

Here,

$f'(x)=2-\dfrac{f(x)}{x}$
or

$\dfrac{dy}{dx}+\dfrac{y}{x}=2$ [ i.e. linear differential equation in

$y$ ]
Integrating Factor,

$IF=e^{\displaystyle \int{\dfrac{1}{x}dx}}=e^{\log x}=x$

$\therefore$ Required solution is

$y.(IF)=\displaystyle \int{Q(IF)dx+C}$

$\Rightarrow y(x)=\int{2(x)dx+C}$

$\Rightarrow yx=x^2+C$

$\therefore y=x+\dfrac{C}{x}$

$[ \because C \neq 0$ , as

$f(1) \neq 1]$
(a)

$\displaystyle \lim_{x \rightarrow 0^+}{f' \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1-Cx^2)}=1$

$\therefore$ Option (a) is correct.
(b)

$\displaystyle \lim_{x \rightarrow 0^+}{xf \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1+Cx^2)}=1$

$\therefore$ Option (b) is correct.
(c)

$\displaystyle \lim_{x \rightarrow 0^+}{x^2f' (x)}=\displaystyle \lim_{x \rightarrow 0^+}{(x^2-C)}=-C\neq 0$

$\therefore$ Option (c) is correct.
(d)

$f(x)=x+\dfrac{C}{x}, C \neq 0$
For

$C > 0, \displaystyle \lim_{x\rightarrow 0^+}{f(x)=\infty}$

$\therefore$ Function is not bounded in

$(0, 2)$ .

$\therefore$ Option (d) is incorrect.

$\displaystyle\lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\cot x-\cos x}{(\pi -2x)^3}$ equals?

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{24}$
0%
$\dfrac{1}{16}$
0%
$\dfrac{1}{8}$

Let

$f : R \to R$ be a differentiable function satisfying

$f'(3) + f'(2) = 0$ .
Then

$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$ is equal to

0%
$e^2$
0%
$e$
0%
$e^{-1}$
0%
$1$

Explanation

$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$

$(1^{\infty}$ form

$)$

$\Rightarrow e^{\underset{x \to 0}{\lim} \dfrac{f(3+x)-f(2-x)-f(3)+f(2)}{x(1+f(2-x)-f(2))}}$

using L'Hopital

$\Rightarrow e^{\underset{x \to 0}{\lim}\dfrac{f'(3+x)+f'(2-x)}{-xf'(2-x)+(1+f(2-x)-f(2))}}$

$\Rightarrow e^{\dfrac{f'(3) + f'(2)}{1}} = 1$

If the function

$f(x)$ satisfies the relation

$f(x+y)=y\dfrac{|x-1|}{(x-1)}f(x)+f(y)$ with

$f(1)=2$ , then

$\displaystyle\lim_{x\rightarrow 1}f'(x)$ is?

0%
$2$
0%
$-2$
0%
$0$
0%
Limit do not exixst

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow a}\dfrac{\sqrt{x}+\sqrt{a}}{x+a}$ .

0%
$-\dfrac{1}{\sqrt{a}}$
0%
$\dfrac{1}{{a}}$
0%
$\dfrac{1}{2\sqrt{a}}$
0%
$\dfrac{1}{\sqrt{a}}$

Explanation

Evaluate

$\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}$

not an indeterminant Form so, limit can be found by simply putting unit of

$x$

$\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}=\dfrac {2\sqrt a}{2a}=\dfrac {1}{\sqrt a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2/3}-9}{x-27}$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{5}$
0%
None of these

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{3x+1}{x+3}$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{5}{3}$
0%
None of these

Explanation

Evaluate

$\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}$

$\to 0, Expression \to \dfrac {1}{3}$ , not indeterminant form

so, Limit can be foord, by simple substituting value

$\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}=\dfrac {1}{3}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}, x > 1$ .

0%
$\dfrac{\sqrt{2}+1}{\sqrt{2}}$
0%
$\dfrac{\sqrt{2}-1}{\sqrt{2}}$
0%
$\dfrac{\sqrt{2}+1}{{2}}$
0%
None of these

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{ax+b}{cx+d}, d\neq 0$ .

0%
$\dfrac{a}{c}$
0%
$\dfrac{a}{d}$
0%
$\dfrac{b}{d}$
0%
None of these

Explanation

$\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}, d\neq 0$

$x\to 0$ , Expression

$\to \dfrac {b}{d}$ , which is a determined Form, so Limit= value

$\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}=\dfrac {b}{a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^2+x^2}-a}{x^2}$ .

0%
$\dfrac{1}{\sqrt a}$
0%
$\dfrac{1}{\sqrt {2a}}$
0%
$\dfrac{1}{a}$
0%
$\dfrac{1}{2a}$

Explanation

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^{2}+x^{2}}-a}{x^{2}}$

$x\rightarrow 0$ , it is

$\dfrac{0}{0}$ Form,

So, using rationalisation

$\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^{2}+x^{2}}-a)}{x^{2}}\times \dfrac{(\sqrt{a^{2}+x^{2}}+a)}{\sqrt{a^{2}+x^{2}}+a)}$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{(a^{2}+x^{2})-a^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}$

$((a+b)(a+b)=a^{2}-b^{2})$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}$

$\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{(\sqrt{a^{2}+x^{2}}+a)}=\dfrac {1}{\sqrt {a^2}+a}=\dfrac{1}{2a}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{2x}{\sqrt{a+x}-\sqrt{a-x}}$ .

0%
$-2\sqrt{a}$
0%
$\sqrt{a}$
0%
$2\sqrt{a}$
0%
None of these

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}$ .

0%
$\dfrac{1}{2\sqrt{a}}$
0%
$\dfrac{1}{2a\sqrt{a}}$
0%
$\dfrac{1}{2a}$
0%
None of these

Explanation

$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a^{2}+ ax}}$

$x \rightarrow 0$ , it is

$\dfrac{0}{0}$ form

So, using rationalization

$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}- \sqrt{a}}{x \sqrt{a^{2}+ax}} \dfrac{(\sqrt{a+x}+ \sqrt{a})}{(\sqrt{a+x}+ \sqrt{a})}$

$\displaystyle \lim_{x \rightarrow 0} \dfrac{x}{x \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}$

$\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{ \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}$

$= \dfrac{1}{2a \sqrt{a}}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 4}\dfrac{2-\sqrt{x}}{4-x}$ .

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
None of these

Explanation

$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{4-\sqrt x}}$

$x\rightarrow 4$ , expression

$\rightarrow \dfrac{0}{0}$ , an indetermined form

using factorisation

$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{2^2-(\sqrt x)^2}}=\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{(2-\sqrt x)(2+\sqrt x)}}=\dfrac{1}{4}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 2}\dfrac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$ .

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$

Explanation

Evaluate

$\displaystyle \lim_{x\to 2}\dfrac {\sqrt {1+4x}-\sqrt {5+2x}}{(x-2)}$

$x\to 2$ , it is

$\dfrac {0}{0}$ From so,

using rationalisation

$\displaystyle \lim_{x\to 2}\dfrac {(\sqrt {1+4x}-\sqrt {5+2x}) (1+4x)+\sqrt {5+2x}}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$

$\displaystyle \lim_{x\to 2}\dfrac {((1+4x)-(5+2x))}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$

$\displaystyle \lim_{x\to 2}\dfrac {2(x-2)}{(x-2)(6)}=\dfrac {1}{3}$

Evaluate the following limits.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}$ .

0%
$\dfrac{1}{\sqrt{2}}$
0%
$-\dfrac{1}{\sqrt{3}}$
0%
$-\dfrac{1}{{2}}$
0%
$-\dfrac{1}{\sqrt{2}}$

Explanation

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}$

$x\rightarrow 0$ , it is

$\dfrac{0}{0}$ from, so,

using rationalisation

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}\dfrac{(\sqrt{2-x}+\sqrt{2+x})}{(\sqrt{2-x}+\sqrt{2+x})}$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x})}}$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{-2x}{x(\sqrt{2-x}+\sqrt{2+x})}}=\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{\sqrt2}$

Evaluate the following limits.
If

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^5-a^5}{x-a}=405$ , find all possible values of a.

0%
$a=3, -3$
0%
$a=2, -2$
0%
$a=5, -5$
0%
None of these

Explanation

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{5}-a^{5}}{x-a}=405$ , find

$a$ ?

Using

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1}$

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{5}-a^{5}}{x-a}=5.a^{4}=405$

$a^{4}=81$

$(a^{2}-9)(a^{2}+9)=0$

$a^{2}-9=0$

$a=3, -3$

Evaluate the following limits.
If

$\displaystyle\lim_{x\rightarrow a}\dfrac{x^9-a^9}{x-a}=9$ , find all possible values of a.

0%
$2, -2$ .
0%
$1, -1$ .
0%
$1, 0$ .
0%
None of these

Explanation

Using formula

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=na^{n-1}$

$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^9-a^9}{x-a}}=9.9^8=9$

$a^8=1\Rightarrow a=\pm 1$

Evaluate the following limit.

$\displaystyle\lim_{x\rightarrow 0}\dfrac{8^x-2^x}{x}$ .

0%
$log 4$
0%
$log 6$
0%
$log 5$
0%
None of these

Explanation

we know

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{a^x-1}{x}}=\log a$

$\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-2^x}{x}}=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1+1-2^x}{x}}$

$=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1}{x}}-\displaystyle \lim_{x\rightarrow 0}{\dfrac{2^x-1}{x}}=\log 8-\log 2$

$=\log 8/2=\log 4$ .

$f : R \rightarrow (0, \infty)$ is an increasing function and if

$\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(3x)}{f(x)} = 1$ , then

$\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(2x)}{f(x)}$ is equal to

0%
$\dfrac{2}{3}$
0%
$\dfrac{3}{2}$
0%
$2$
0%
$3$
0%
$1$

Explanation

Given

$f : R \rightarrow (0, \infty)$ is an increasing function.

And

$\underset{x \rightarrow 2018}{\lim} \dfrac{f(3x)}{f(x)} = 1$

So,

$\underset{x \rightarrow 2018}{\lim} f(3x) = \underset{x \rightarrow 2018}{\lim} f(x)$

$\Rightarrow f(x) =$ constant.

Therefore,

$\underset{x \rightarrow 2018}{\lim} \dfrac{f(2x)}{f(x)} = 1$

$\underset{x\to 0}{\lim} \left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2}$ is equal to:

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$\dfrac{1}{e^2}$
0%
$\dfrac{1}{e}$
0%
$e^2$
0%
$e$

Explanation

$L = \underset{x \to 0}{\lim} \left(\dfrac{3x^2 + 2}{7x^2 + 2}\right)^{\tfrac{1}{x^2}}$

As it is a

$1^{\infty}$ form

We know the formula,

$\Rightarrow$

$\displaystyle\lim_{x \to 0} f(x)^{g(x)}= \displaystyle e^{\underset{x\to 0}{\lim} g(x) \left (f(x)-1\right)}$

$= e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2}{7x^2 + 2}\displaystyle-1\right)}$

$= e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2-7x^2-2}{7x^2 + 2}\right)}$

$= e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{-4x^2}{7x^2 + 2}\right)}$

$=e^{\tfrac{-4}{2}}$

$=e^{-2}$

$=\dfrac1{e^{2}}$

$\boxed{L=\dfrac1{e^{2}}}....Answer$

Hence option

$'A'$ is the answer.

$f$ is differentiable at

$x = 1$ and

$\underset{h \rightarrow 0}{\lim} \dfrac{1}{h} f (1 + h) = 5, f'(1) =$

0%
$0$
0%
$1$
0%
$3$
0%
$4$
0%
$5$

Explanation

$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)-f(1)}{h}$ ; function is differentiable.

and

$\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$ [Given function is continuous]

$\Rightarrow f(1)=0$

Hence,

$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$

The value of

$\displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}$ is

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1/3
0%
2/3
0%
-1/4
0%
3/2

Explanation

We have

$\displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}$

$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{(1+\cos x)\left(1-\cos x+\cos ^{2} x\right)}{(1-\cos x)(1+\cos x)}$

$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{1-\cos x+\cos ^{2} x}{1-\cos x}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}$ is equal to

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0
0%
$\infty$
0%
-2
0%
2

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \dfrac{x}{2}}$

$=2 \displaystyle \lim _{x \rightarrow 0}\left[\dfrac{(x / 2)^{2}}{\sin ^{2} \dfrac{x}{2}}\right]\left(\dfrac{e^{x}-1}{x}\right)=2$

$\displaystyle \lim _{x \to \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right]$ is equal to

0%
1
0%
-1
0%
0
0%
$None \ of \ these$

Explanation

$\displaystyle \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right]$

$=\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{2 x \sin x-\pi}{2 \cos x} \quad\quad\quad \left(\dfrac{0}{0}form\right)$

$=\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{[2 \sin x+2 x \cos x]}{-2 \sin x}$
(Applying L'Hopital's rule)

$=-1$

$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}$ is equal to

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$-\dfrac{1}{2 \sqrt{2}}$
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$\dfrac{1}{2 \sqrt{2}}$
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$\dfrac{1}{\sqrt{2}}$
0%
Does not exist

Explanation

$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}= \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{-x \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}$

$=-\displaystyle \lim _{x \rightarrow-\infty} \dfrac{\tan \dfrac{1}{x}}{\dfrac{1}{x} \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}=-\dfrac{1}{2 \sqrt{2}}$

$\displaystyle\lim_{n\rightarrow \infty}\dfrac{(n 1)^{1/n}}{n}$ equals?

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$1$
0%
$e$
0%
$e^{-1}$
0%
None of these

$f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}},$ then

0%
$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty }$
0%
$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty }$
0%
$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty }$
0%
none of these

Explanation

$\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \dfrac{-\sin t}{(1-\cos t)^{1 / 3}}$

$=-\lim _{t \rightarrow 0} \dfrac{2 \sin \dfrac{t}{2} \cos \dfrac{t}{2}}{\left(2 \sin ^{2} \dfrac{t}{2}\right)^{1 / 3}}$

$=-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \dfrac{t}{2}\left(\sin \dfrac{t}{2}\right)^{1 / 3}=0$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n)$ is equal to

0%
1
0%
0
0%
n/m
0%
None of these

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}}=\lim _{x \rightarrow 0}\left(\dfrac{\sin x^{n}}{x^{n}}\right)\left(\dfrac{x^{n}}{x^{m}}\right)\left(\dfrac{x}{\sin x}\right)^{m}$

$=\displaystyle \lim _{x \rightarrow 0} x^{n-m}=0 \quad[\because m<n]$

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x}$ is equal to

0%
0
0%
1
0%
2
0%
4

Explanation

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x}$

$=\displaystyle \lim _{x \rightarrow 1} \dfrac{1-\cos \left(\dfrac{3 \pi}{2}-\dfrac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)}$

$=\lim _{x \rightarrow 1} \dfrac{2 \sin ^{2}\left(\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}\right)}{2 \sin ^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}$

$=\displaystyle \lim _{x \rightarrow 1}\left(\dfrac{\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}}{\dfrac{\pi}{2}-\dfrac{\pi x}{2}}\right)^{2}$

$=\displaystyle \lim _{x \rightarrow 1} 9\left(\dfrac{\dfrac{1}{2}-\dfrac{x}{1+x^{2}}}{1-x}\right)^{2}=\lim _{x \rightarrow 1} 9\left(\dfrac{x-1}{2\left(1+x^{2}\right)}\right)^{2}=0$

The value of

$\displaystyle \lim _{x \rightarrow 2} \dfrac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}$ is

0%
16
0%
8
0%
4
0%
2

Explanation

$\displaystyle \lim _{x \rightarrow 2} \dfrac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}$

$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}\right)^{2}-6 \times 2^{x}+2^{3}}{\sqrt{2^{x}}-2}\left[\text { Multiplying } N^{r} \text { and } D^{r} \text { by } 2^{x}\right]$

$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}-4\right)\left(2^{x}-2\right)(\sqrt{2^{x}}+2)}{(\sqrt{2^{x}}-2)(\sqrt{2^{x}}+2)}$

$=\displaystyle \lim _{x \rightarrow 2} \dfrac{\left(2^{x}-4\right)\left(2^{x}-2\right)(\sqrt{2^{x}}+2)}{\left(2^{x}-4\right)}$

$=\displaystyle \lim _{x \rightarrow 2}\left(2^{x}-2\right)(\sqrt{2^{x}}+2)=\left(2^{2}-2\right)(2+2)=8$

$The \ value \ of \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}}$ is

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$e^{-2 \pi}$
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$e^{1 / \pi}$
0%
$e^{2 /\pi}$
0%
$e^{-1 / \pi}$

Explanation

$\displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}}$

$=\displaystyle \lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \dfrac{\pi x}{2}}$

$=e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}}$

$=e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \cot \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}$

$=e^{\displaystyle \lim _{x \rightarrow 1} \dfrac{(1-x)}{\tan \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}}$

$=e^{\dfrac{2}{\pi}\displaystyle \lim_{x \to 1}{\dfrac{\dfrac{\pi}{2}(1-x)}{tan \left(\dfrac{\pi}{2}(1-x)\right)}}}$

$=e^{\dfrac{2}{\pi}}$

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x} \text { is equal to }$

0%
$\dfrac{1}{2 \pi}$
0%
$\dfrac{-1}{\pi}$
0%
$\dfrac{-2}{\pi}$
0%
None of these

Explanation

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x}$

$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)}$

$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \dfrac{1+x}{2 \pi}=\dfrac{-1}{\pi}$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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