Limits and Continuity Test 9

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Limits and Continuity Test 9

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Question 1
1 / -0

$$\displaystyle \lim_{n\rightarrow \infty }\left(\displaystyle \frac{e^{n}}{\pi}\right)^{1/n}=$$

A
0

B
1

C
$$\mathrm{e}^{2}$$

D
$$\mathrm{e}$$

Solution

Let $$y$$ = $$\displaystyle\lim _{ n\rightarrow \infty }{ { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{\displaystyle \frac { 1 }{ n } } }$$

Applying logarithm on both sides

$$\log { y } $$ = $$\displaystyle\lim _{ n\rightarrow \infty }{ \log { { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{ {\displaystyle \frac { 1 }{ n } } } } } =\quad \lim _{ n\rightarrow \infty }{\displaystyle \frac { 1 }{ n } \log { { \left( \displaystyle\frac { e^{ n } }{ \pi } \right) } } }$$

$$\log { y } $$ = $$\lim _{ n\rightarrow \infty }{\displaystyle \frac { 1 }{ n } \left( n\log e-\log { \pi } \right) } $$

By L'Hospital Rule

$$\log y=\log e$$

$$ \Rightarrow \quad y = e$$

$$\therefore \quad \displaystyle\lim _{ n\rightarrow \infty }{ { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{\displaystyle \frac { 1 }{ n } } } = e$$
Question 2
1 / -0

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

A
$$e^{\displaystyle \frac{1}{\pi}}$$

B
$$e^{\displaystyle \frac{2}{\pi}}$$

C
$$-e^{\displaystyle \frac{2}{\pi}}$$

D
$$\mathrm{e}$$

Solution

Let $$k=\lim _{ x\rightarrow 1 } (2-x)^{ \tan \left(\dfrac { \pi x }{ 2 } \right) }$$
$$\log { k } =\lim _{ x\rightarrow 1 } tan\left( \dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } $$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } tan\left( \displaystyle\dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } }$$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { \log { \left( 2-x \right) } }{ \cot { \left( \dfrac { \pi x }{ 2 } \right) } } }$$
$$k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { -1 }{ -(2-x)\csc ^{ 2 }{ \left( \dfrac { \pi x }{ 2 } \right) \left( \dfrac { \pi }{ 2 } \right) } } }$$
$$\Rightarrow k={e}^{\displaystyle\dfrac{2}{\pi}}$$
Question 3
1 / -0

lf the function defined by $$\mathrm{f}({x})=\displaystyle \frac{\sin 3(x-p)}{\sin 2(x-p)}$$ for $${x}\neq {p}$$ is continuous at $${x}={p}$$ then $$\mathrm{f}({p})=$$

A
$$\dfrac{3}{2}$$

B
$$\dfrac{2}{3}$$

C
$$6$$

D
$$\dfrac{1}{6}$$

Solution

Since, it is given the function is continuous at $$x = p,$$we calculate
$$\lim _{ x \rightarrow p } \dfrac{sin 3(x-p)}{sin 2(x-p)} $$
$$= lim _{ x \rightarrow p } \dfrac{sin 3(x-p)}{3(x-p)} \times \dfrac{2(x-p)}{sin 2(x-p)} \times \dfrac{3}{2} $$
Using the property of limits,
$$f(p) = 1 \times 1 \times \dfrac{3}{2} $$
Question 4
1 / -0

If the function $$\displaystyle \mathrm{f}({x})=\begin{cases}\dfrac{2^{x+2}-16}{4^{x}-16}&& for {x}\neq 2\\ \mathrm{A} && x =2\end{cases}$$ is continuous at $$x =2$$, then $$\mathrm{A}=$$

A
$$2$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{1}{4}$$

D
$$0$$

Solution

Since, the function is continuous at $$x = 2$$
$$A = \displaystyle\lim _{x \rightarrow 2 } f(x) $$
Since, on applying the limit this is of the $$ \dfrac{0}{0} $$ form, we apply L-Hospital's Rule
$$\Rightarrow\displaystyle\lim_{ x \rightarrow 2}\dfrac{4.{2}^{x}log 2}{{4}^{x} log 4} $$
Now applying the limit,
$$A = \dfrac{1}{2} $$
Question 5
1 / -0

$$f(x)=x\left[3-\displaystyle \log\left(\frac{\sin {x}}{x}\right)\right]-2$$ to be continuous at $${x}=0$$, then $$\mathrm{f}({0})=$$

A
$$0$$

B
$$2$$

C
$$-2$$

D
$$3$$

Solution

USing the property of limits, $$\underset{ x \rightarrow 0 }{Lim} \dfrac{sin x}{x} = 1 $$
Thus, applying the limit to the expression,
$$\underset{ x \rightarrow 0 }{Lim} \ \ x\left[3 - log\left(\dfrac{sin x}{x}\right)\right] -2 $$
$$= 0(3 - log 1) - 2$$
$$= 0 - 2$$
$$ = -2 $$
Question 6
1 / -0

Let $$\displaystyle \mathrm{f}({x})=\begin{cases} \dfrac{(e^{kx}-1).\sin kx}{x^{2}} & for \ {x}\neq 0 \\ 4 & for \ {x} =0\end{cases}$$ is continuous at $${x}=0$$ then $${k}=$$

A
$$\pm 1$$

B
$$\pm 2$$

C
0

D
$$\pm 3$$

Solution

For a given function to be continuous at $$x=0$$

$$\displaystyle \lim_{x\to 0} = f(0)\Rightarrow \lim_{x\to 0}\frac{(e^{kx}-1).\sin kx}{x^2} =4$$

$$\Rightarrow \displaystyle k^2. \lim_{x\to 0}\frac{e^{kx}-1}{kx}.\frac{\sin kx}{kx} =4$$

$$\Rightarrow k^2 = 4\Rightarrow k = \pm 2$$
Question 7
1 / -0

lf $$f(x)=
\left\{\begin{matrix} (1+|\sin x|)^{\displaystyle \frac{a}{|\sin x|}}&-\displaystyle \frac{\pi}{6}<x<0\\ b&x=0 \\ e^{\displaystyle \frac{\tan 2x}{\tan 3x}} &0<x<\displaystyle \frac{\pi}{6}\end{matrix}\right.$$ is

continuous at $$\mathrm{x}=0$$ then

A
$$a=e^{2/3},b=\dfrac{2}{3}$$

B
$$a=\dfrac{2}{3},b=e^{2/3}$$

C
$$a=\dfrac{1}{3},b=e^{1/3}$$

D
$$a=e^{1/3},b=e^{1/3}$$

Solution

For the function $$f(x)$$ to be continuous at $$x = 0$$
$$\lim_{ x \rightarrow {0}^{-}} f(x) = f(0) = \lim_{ x \rightarrow {0}^{+}} f(x) $$
LHL
$$\lim_{ x \rightarrow {0}^{-}} exp( |\sin x| \times \dfrac{a}{|\sin x|}) = e^a$$
$$f(0) = b$$
RHL
$$\lim_{ x \rightarrow {0}^{+} } exp(\dfrac{\tan 2x}{\tan 3x}) $$
$$= \lim_{ x \rightarrow {0}^{+}} exp(\dfrac{\tan 2x}{2x\ tan3x} \times {3x} \times \dfrac{2}{3}) = e ^{\frac{2}{3}} $$
Thus,
$$exp(a) = b = exp( \dfrac{2}{3} )$$
Hence, $$a = \dfrac{2}{3} $$ and $$b = exp( \dfrac{2}{3} )$$
Question 8
1 / -0

lf the function $$f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x}, & x\neq\dfrac{\pi}{2}\\ 3 & at x=\dfrac{\pi}{2}\end{cases}$$is continuous at $$\displaystyle {x}=\dfrac{\pi}{2}$$ then $${k}=$$

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

On applying the limits, since the function is of the form $$ \dfrac{0}{0} $$, We apply L-Hospital's rule,
$$\displaystyle\lim_{ x \rightarrow \dfrac{\pi}{2}} \dfrac{-ksin x}{-2} $$
Since, the function has to be continuous at $$x = \dfrac{\pi}{2} $$
$$ \Rightarrow\dfrac{k}{2} = 3$$
$$\Rightarrow k = 6$$
Question 9
1 / -0

The function $$f(x)=\begin{cases} 0,& \text{x is irrational }\\ 1,& \text{x is rational }\end{cases}$$ is

A
continuous at $$x=1$$

B
discontinuous only at $$0$$

C
discontinuous only at 0,1

D
discontinuous everywhere

Solution

From the number theory, we already know that between any 2 rational numbers there exists an irrational number and vice versa.
Thus, for the function f(x) as defined above it will take both the values 0 and 1 in the neighbourhood of every point x = a.
Thus, function can never be continuous.
Question 10
1 / -0

The function $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\cos x-\sin x}{\cos 2x}$$ is not defined at $$x=\displaystyle \frac{\pi}{4}$$ The value of $$f\left(\displaystyle \frac{\pi}{4}\right)$$ so that $$\mathrm{f}(\mathrm{x})$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ is

A
$$\displaystyle \frac{1}{\sqrt{2}}$$

B
$$\sqrt{2}$$

C
$$-\sqrt{2}$$

D
$$1$$

Solution

It is given that the function is not defined at x = $$ \dfrac{\pi}{4} $$
So, $$f\left( \dfrac{\pi}{4} \right)$$ =$$\underset{x \rightarrow \dfrac{\pi}{4}}{Lim}\ \dfrac{cos x - sin x}{cos 2x} $$
Since, this is of the $$ \dfrac{0}{0} $$ form, we apply L-Hospital's Rule,
$$\underset{x \rightarrow \dfrac{\pi}{4}}{Lim} \ \ \dfrac{-sin x - cos x}{-2sin 2x} $$
Now, applying the limit,
$$f\left( \dfrac{\pi}{4}\right) $$ = $$ \dfrac{1}{{2}^{0.5}} $$

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 9

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Limits and Continuity Test 9

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SHARING IS CARING

Question 1

0

1

$$\mathrm{e}^{2}$$

$$\mathrm{e}$$

Solution

Question 2

$$e^{\displaystyle \frac{1}{\pi}}$$

$$e^{\displaystyle \frac{2}{\pi}}$$

$$-e^{\displaystyle \frac{2}{\pi}}$$

$$\mathrm{e}$$

Solution

Question 3

$$\dfrac{3}{2}$$

$$\dfrac{2}{3}$$

$$6$$

$$\dfrac{1}{6}$$

Solution

Question 4

$$2$$

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

$$\dfrac{1}{4}$$

$$0$$

Solution

Question 5

$$0$$

$$2$$

$$-2$$

$$3$$

Solution

Question 6

$$\pm 1$$

$$\pm 2$$

0

$$\pm 3$$

Solution

Question 7

$$a=e^{2/3},b=\dfrac{2}{3}$$

$$a=\dfrac{2}{3},b=e^{2/3}$$

$$a=\dfrac{1}{3},b=e^{1/3}$$

$$a=e^{1/3},b=e^{1/3}$$

Solution

Question 8

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 9

continuous at $$x=1$$

discontinuous only at $$0$$

discontinuous only at 0,1

discontinuous everywhere

Solution

Question 10

$$\displaystyle \frac{1}{\sqrt{2}}$$

$$\sqrt{2}$$

$$-\sqrt{2}$$

$$1$$

Solution

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