Limits and Continuity Test 31

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Limits and Continuity Test 31

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Weekly Quiz Competition

Question 1
1 / -0

$$ \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10) $$ is equal to

A
0

B
1

C
19

D
20

Solution

$$ \because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$$
Here, for $$ x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1 $$
and in all other cases it is zero.
$$ \therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1 $$
Question 2
1 / -0

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)} $$ is equal to

A
2

B
-2

C
1

D
-1

Solution

$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$$
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$$
$$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$$
Question 3
1 / -0

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} $$ is equal to

A
2

B
-2

C
1/2

D
-1/2

Solution

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$$
$$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$$
$$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$$
Question 4
1 / -0

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$$ is equal to

A
$$e^{\sin ^{2} y}$$

B
$$\sin 2 y e^{\sin ^{2} y}$$

C
0

D
None of these

Solution

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]=\lim _{x \rightarrow 0} \dfrac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t \\$$

$$\therefore\left(\dfrac{0}{0} \text { form }\right)\\$$

Apply L'Hopital Rule
$$\displaystyle =\lim _{x \rightarrow 0} \dfrac{e^{\sin ^{2}(x+y)}\left(1+\dfrac{d y}{d x}\right)-e^{\sin ^{2} y} \dfrac{d y}{d x}}{1} \\$$

$$=e^{\sin ^{2} y}\left[1+\dfrac{d y}{d x}-\dfrac{d y}{d x}\right]=e^{\sin ^{2} y}$$
Question 5
1 / -0

If $$y^{r}=\dfrac{n !^{n+r-1} C_{r-1}}{r^{n}},$$ where $$n=k r(k \text { is constant }),$$ then $$\operatorname{lim}_{r\rightarrow\infty} y$$ is equal to

A
$$(k-1) \log _{e}(1+k)-k$$

B
$$(k+1) \log _{e}(k-1)+k$$

C
$$(k+1) \log _{e}(k-1)-k$$

D
$$(k-1) \log _{e}(k-1)+k$$

Solution

$$y^{r}=\left(1+\dfrac{1}{r}\right)\left(1+\dfrac{2}{r}\right)\left(1+\dfrac{3}{r}\right) \cdots\left(1+\dfrac{n-1}{r}\right)\\$$
$$\Rightarrow \log y=\dfrac{1}{r} \sum_{p=1}^{n-1} \log \left(1+\dfrac{p}{r}\right)\\$$
$$\Rightarrow \lim _{n \rightarrow \infty} y=\lim _{r \rightarrow \infty} y=\int_{-\infty}^{k} \log (1+x) d x=(k-1) \log _{e}(1+k)-k$$
Question 6
1 / -0

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}} $$ is equal to

A
1/6

B
-1/3

C
1/2

D
1

Solution

$$ \cos (\tan x)-\cos x=2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)$$
$$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}}=\lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}}$$
$$ =\displaystyle \lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}\left(\dfrac{x+\tan x}{2}\right)\left(\dfrac{x-\tan x}{2}\right)}\left(\dfrac{x^{2}-\tan ^{2} x}{4}\right) $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\tan ^{2} x}{x^{4}} $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\left(x+\dfrac{x^{3}}{3}+\dfrac{2}{15} x^{5}+\cdots\right)^{2}}{x^{4}} $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x^{2}}\left(1-\left(1+\dfrac{x^{2}}{3}+\dfrac{2}{15} x^{4}+\dots\right)^{2}\right)=-\dfrac{1}{3} $$
Question 7
1 / -0

The value of $$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}} $$ is

A
$$\dfrac{2 a}{\pi} $$

B
$$-\dfrac{2 a}{\pi} $$

C
$$ \dfrac{4 a}{\pi} $$

D
$$-\dfrac{4 a}{\pi} $$

Solution

$$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$$
$$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$$
$$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$$
Question 8
1 / -0

The value of $$ \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\ldots .+\dfrac{e^{(n-1) / n}}{n}\right] $$ is

A
1

B
0

C
e-1

D
e+1

Solution

$$ \quad \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\dots+\dfrac{e^{(n-1) / n}}{n}\right] $$
$$=\displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1+e^{1 / n}+\left(e^{1 / n}\right)^{2}+\cdots+\left(e^{1 / n}\right)^{n-1}}{n}\right] $$
$$=\displaystyle \lim _{n \rightarrow \infty} \dfrac{1 \cdot\left[\left(e^{1 / n}\right)^{n}-1\right]}{n\left(e^{1 / n}-1\right)}=(e-1) \lim _{n \rightarrow \infty} \dfrac{1}{\left(\dfrac{e^{1 / n}-1}{1 / n}\right)} $$
$$=(e-1) \times 1=(e-1) $$
Question 9
1 / -0

If function $$f(x)=\dfrac{x^2-9}{x-3}$$ is continuous at $$x=3$$, then value of $$(3)$$ will be:

A
$$6$$

B
$$3$$

C
$$1$$

D
$$0$$

Solution

$$f(x)=\dfrac{x^2-9}{x-3}$$
Left hand limit
$$f(3+0)=\displaystyle \lim_{h \rightarrow 0}f(3+h)$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{(3+h)^2-9}{3+h-3}$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{h(6+h)}{h}$$
$$=\displaystyle \lim_{h \rightarrow 0} (6+h)$$
$$=6$$
Function is continous at $$x=3$$, so
$$f(3)=f(3+0)$$
$$f(3)=6$$
Hence, option $$(a)$$ is correct.
Question 10
1 / -0

If $$f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
is continuous at $$x=0$$ then the value of $$k$$ will be:

A
$$0$$

B
$$m+n$$

C
$$m-n$$

D
$$m.n$$

Solution

$$\therefore$$ Function is continous at $$x=0$$
$$\therefore \displaystyle \lim_{x \rightarrow 0} f(x)=f(0)$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{\log (1+mx)- \log (1-nx)}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{ [mx-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3} - ...]- [nx-\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} - ...]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{x \left[ m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... \right]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... =k$$
$$\Rightarrow m+n=k$$
$$\Rightarrow k=m+n$$
Hence, option $$(b)$$ is correct.

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 31

Result

Limits and Continuity Test 31

Score

-

Rank

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SHARING IS CARING

Question 1

0

1

19

20

Solution

Question 2

2

-2

1

-1

Solution

Question 3

2

-2

1/2

-1/2

Solution

Question 4

$$e^{\sin ^{2} y}$$

$$\sin 2 y e^{\sin ^{2} y}$$

0

None of these

Solution

Question 5

$$(k-1) \log _{e}(1+k)-k$$

$$(k+1) \log _{e}(k-1)+k$$

$$(k+1) \log _{e}(k-1)-k$$

$$(k-1) \log _{e}(k-1)+k$$

Solution

Question 6

1/6

-1/3

1/2

1

Solution

Question 7

$$\dfrac{2 a}{\pi} $$

$$-\dfrac{2 a}{\pi} $$

$$ \dfrac{4 a}{\pi} $$

$$-\dfrac{4 a}{\pi} $$

Solution

Question 8

1

0

e-1

e+1

Solution

Question 9

$$6$$

$$3$$

$$1$$

$$0$$

Solution

Question 10

$$0$$

$$m+n$$

$$m-n$$

$$m.n$$

Solution

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