Limits and Continuity Test 26

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Limits and Continuity Test 26

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Question 1
1 / -0

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

A
0

B
1

C
2

D
DNE

Solution
Question 2
1 / -0

Let $$U_{ { n } }=\dfrac { n! }{ (n+2)! } $$ where $$n \in N .$$ If $$S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }$$ then $$\lim _ { n \rightarrow \infty } \mathrm { S } _ { n }$$ equals :

A
$$2$$

B
$$1$$

C
$$1/2$$

D
$$1/3$$

Solution

$$\begin{array}{l} { U_{ n } }=\frac { { n! } }{ { \left( { n+2 } \right) ! } } =\frac { 1 }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } =\frac { { \left( { n+1 } \right) -\left( { n+1 } \right) } }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } \\ =\frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } \\ { S_{ n } }=\sum _{ n=1 }^{ n }{ { U_{ n } } } =\left[ { \left( { \frac { 1 }{ 2 } -\frac { 1 }{ 3 } } \right) +\left( { \frac { 1 }{ 3 } -\frac { 1 }{ 4 } } \right) +............+\left( { \frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } } \right) } \right] \\ { S_{ n } }=\frac { 1 }{ 2 } -\frac { 1 }{ { n+2 } } \\ { \lim }_{ n\to \infty }{ S_{ n } }=\frac { 1 }{ 2 } \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 3
1 / -0

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

A
$$4$$

B
$$2$$

C
$$1$$

D
$$\dfrac{1}{2}$$

Solution

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.
Question 4
1 / -0

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

A
$$a = - 1 , b = 1 / 2$$

B
$$a = 1 , b = 1 / 2$$

C
$$a = 1 , b = - 1 / 2$$

D
None of these

Solution

We have,
$$\begin{array}{l} { \lim }_{ x\to \infty }\left( { \sqrt { { x^{ 2 } }-x+1 } -ax-b } \right) =0 \\ Put\, \, x=\cfrac { 1 }{ t } \\ { \lim }_{ t\to { 0^{ + } } }\left( { \sqrt { \cfrac { 1 }{ { { t^{ 2 } } } } -\cfrac { 1 }{ t } +1 } -\cfrac { a }{ t } -6 } \right) =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { { { \left( { 1-t+{ t^{ 2 } } } \right) }^{ \cfrac { 1 }{ 2 } } }-a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-\cfrac { t }{ 2 } +\cfrac { { { t^{ 2 } } } }{ 2 } } \right) -a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-a } \right) +t\left( { -b-\cfrac { 1 }{ 2 } } \right) } }{ t } =0 \\ 1=a \\ and\, \, b=-\cfrac { 1 }{ 2 } \\ Option\, \, \, C\, \, \, is\, \, correct\, \, answer. \end{array}$$
Question 5
1 / -0

The value of $$\lim_{x\rightarrow \infty }$$ y In $$(\frac{sin (x+1/y)}{sin x})$$ when $$0 < x < \pi /2$$ is

A
cos x

B
$$\infty$$

C
cot x

D
does not exist
Question 6
1 / -0

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

A
$$\dfrac{k}{e}$$

B
$$\dfrac{e}{k}$$

C
$$\dfrac{1}{ke}$$

D
None of these

Solution
Question 7
1 / -0

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

A
$$f(x)$$ is continuous but not differentiable at $$x=0$$

B
$$f(x)$$ is both continuous and differentiable at $$x=0$$

C
$$f(x)$$ is neither continuous nor differentiable at $$x=0$$

D
$$f(x)$$ is a periodic function

Solution
Question 8
1 / -0

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

A
$$\pi/4$$

B
$$\tan^{-1}{(2)}$$

C
$$\pi/2$$

D
$$\tan^{-1}{(3)}$$

Solution
Question 9
1 / -0

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

A
$$1$$

B
$$0$$

C
$$\infty$$

D
Does not exist

Solution
Question 10
1 / -0

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

A
$$8$$

B
$$8\sqrt { 2 } $$

C
$$4$$

D
$$4\sqrt { 2 } $$

Solution

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