Limits and Continuity Test 13

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Limits and Continuity Test 13

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Question 1
1 / -0

The value of $$\mathrm{f}({0})$$ so that the function $$\displaystyle \mathrm{f}({x})=\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$$ becomes continuous is equal to

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

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

C
$$2$$

D
$$\dfrac{1}{3}$$

Solution

Since, f is continuous
$$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { \sqrt { 1+x } -\sqrt [ 3 ]{ 1+x } }{ x } } $$
It is of $$\dfrac{0}{0}$$ form
By L-Hospital's rule
$$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \left( \frac { 1 }{ 2\sqrt { 1+x } } -\frac { 1 }{ 3{ \left( 1+x \right) }^{ 2/3 } } \right) } =\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } $$
Question 2
1 / -0

If $$f(x)$$ is continuous and $$f\left(\dfrac {9}{2}\right)=\dfrac {2}{9}$$, then $$\displaystyle\lim _{ x\rightarrow 0 }f\left(\frac {1-\cos 3x}{x^2}\right)$$ is equal to :

A
$$\dfrac {9}{2}$$

B
$$\dfrac {2}{9}$$

C
0

D
$$\dfrac {8}{9}$$

Solution

$$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { 1-cos3x }{ x^{ 2 } } }$$

$$ =\lim _{ x\rightarrow 0 }{ \dfrac { 2\sin ^{ 2 }{ (\frac { 3x }{ 2 } ) } }{ (\frac { 3x }{ 2 } )^{ 2 } } } \times \dfrac { 9 }{ 4 } $$

$$\displaystyle =\frac{9}{2}$$

$$\Rightarrow \displaystyle \lim _{ x\rightarrow 0 }{ f\left( \frac { 1-cos3x }{ x^{ 2 } } \right) } = f\left( \dfrac 9 2\right)=\dfrac { 2}{ 9 } $$
.
Question 3
1 / -0

If $$\displaystyle f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $$ then $$\mathop {\lim }\limits_{x \to \infty } f(x)$$ is

A
O

B
$$\infty$$

C
1

D
None of these

Solution

$$\displaystyle f(x) =\sqrt{\frac{x-sin x}{x+ cos^2x}} =\sqrt{\frac{1- \frac{sin x}{x}}{1+ \frac{cos^2x}{x}}}$$
$$\lim_{x \rightarrow \infty}f(x)=1$$ $$\displaystyle \lim_{x \rightarrow \infty} \frac{sin x}{x}=0$$
$$\displaystyle \lim_{x \rightarrow \infty} \frac{cos^2 x}{x}=0$$
Question 4
1 / -0

$$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$$ is

A
$$\ln (n!)$$

B
$$n$$

C
$$n!$$

D
$$0$$

Solution

Consider that $$y=\displaystyle\lim_{x\to\infty}(1^x+2^x+3^x+.........+n^x)^{1/x}$$
take natural logarithm on both sides, we have
$$\ln y=\displaystyle\lim_{x\to\infty}\ln\left[(1^x+2^x+3^x+.........+n^x)^{1/x}\right]$$
$$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{\ln (1^x+2^x+3^x+.........+n^x)}{x}$$
RHS has $$\dfrac{\infty}{\infty}$$ form. Hence, use L'hospital's rule
$$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n }{1^x+2^x+3^x+.........+n^x}$$
On approximating this form, we have
$$\ln y=\displaystyle\lim_{x\to\infty}\dfrac{n^x\ln n }{n^x}$$
$$\ln y=\ln n$$
$$\therefore y=n$$
Hence, this is required answer.
Question 5
1 / -0

$$\displaystyle \lim_{x\to\infty}{\displaystyle \frac{(2x + 1)^{40}(4x - 1)^5}{(2x + 3)^{45}}}$$ is equal to

A
16

B
24

C
32

D
8

Solution

$$\displaystyle\lim_{x\to\infty}\frac{(2x+1)^{40}(4x-1)^5 }{(2x+3)^{45}}$$
$$=\displaystyle\lim_{x\to\infty}\frac{\left(2+\Large \frac{1}{x}\right)^{40}\left(4-\Large \frac{1}{x}\right)^5}{\left(2+\Large \frac{3}{x}\right)^{45}}$$$$=\displaystyle \frac{2^{40}4^5}{2^{45}}\\$$
$$=2^5 = 32$$
Question 6
1 / -0

If $$f(x) =\left\{\begin{matrix}
\dfrac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}},x>0 & \\
e^{x}\sin x+\pi x+\lambda \ln 4,x\leq 0 &
\end{matrix}\right.$$is continuous at $$x= 0$$, then $$\lambda$$ is a

A
rational number

B
irrational number

C
natural number

D
complex number

Solution

$$\displaystyle\lim_{x\to-h} f(x) =\lambda \ln4$$
$$\displaystyle\lim_{x\to 0+h} f(x)= \dfrac { 2^{ 3x }-2^{ 2x }-2^{ x }+1^{ x } }{ x^{ 2 } } =\lambda \ln4$$

$$\displaystyle\Rightarrow \lim_{h\to 0}\dfrac { 2^{ 3h }-2^{ 2h }-2^{ h }+1^{ h } }{ h^{ 2 } } =\lambda \ln4$$

$$\displaystyle\Rightarrow \lim_{ h\to 0}\dfrac{(2^{ 2h }(2^{ h }-1)-1(2^{ h }-1))}{h^{ 2 }}=\lambda \ln4$$

$$\displaystyle\Rightarrow \lim_{h\to 0}\dfrac{(2^{ h }+1){ (2^{ h }-1) }^{ 2 }}{h^{ 2} }=\lambda \ln4$$

$$\displaystyle\Rightarrow \lim_{ h\to 0} \dfrac{2{ (2^{ h }-1) }^{ 2 }}{h^{ 2 }}=\lambda \ln4$$

Applying L'hospital's rule
$$\displaystyle\Rightarrow \lim_{h\to 0}(2^{ h }-1)\dfrac{\ln4}{h} =\lambda \ln4$$
$$\Rightarrow \ln2\ln4 = \lambda \ln4$$
$$\Rightarrow \ln2=\lambda$$

Hence, option 'B' is correct.
Question 7
1 / -0

The function $$f :( R-{0})$$ $$\rightarrow $$ R given by $$\displaystyle f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}$$ can be made continuous at $$x = 0$$ by defining $$f(0)$$ as

A
$$2$$

B
$$-1$$

C
$$0$$

D
$$1$$

Solution

Given$$\displaystyle f\left( x \right) =\frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } $$
$$ \displaystyle \Rightarrow f\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \left\{ \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } \right\} } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 2x }-1-2x }{ x\left( { e }^{ 2x }-1 \right) } } \ ....... \quad \left[ \frac { 0 }{ 0 } form \right] $$
$$\therefore $$ usingL'Hospital rule
$$f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) } } $$
$$\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } } } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right) } =1$$
Question 8
1 / -0

$$\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n + 1)^2}{(n + 2)(n^2 + 3n - 1)}}$$ is equal to

A
$$0$$

B
$$2$$

C
$$4$$

D
$$\infty$$

Solution

$$\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n+1)^2}{(n+2)(n^2+3n-1)} }$$

$$ = \displaystyle \lim_{n\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{n} \right)^2}{\left(1+\Large \frac{2}{n} \right)\left(1+\Large \frac{3}{n} - \Large \frac{1}{n^2} \right)} }$$

$$= \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)} = 4$$
Question 9
1 / -0

Let $$\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R}$$ be defined by $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\mathrm{k}-2\mathrm{x}, \mathrm{i}\mathrm{f} \mathrm{x}\leq-1\\
2\mathrm{x}+3, \mathrm{i}\mathrm{f} \mathrm{x}>-1
\end{array}\right.$$ be continous. then find possible value of $$\mathrm{k}$$ is

A
$$0$$

B
$$\displaystyle -\frac{1}{2}$$

C
$$-1$$

D
$$1$$

Solution

$$\mathrm{f}(\mathrm{x})=\mathrm{k}-2\mathrm{x}$$ if $$\mathrm{x}\leq-1$$

$$=2\mathrm{x}+3$$ if $$\mathrm{x}>-1$$

So, $$L.H.L.= R.H.L.$$

$$k-2x=2x+3$$

$$k=4x+3$$

For $$x=-1$$, $$k=-1$$

Hence, option $$'C'$$ is correct.
Question 10
1 / -0

If $$f(x) = \displaystyle \left\{\begin{matrix}x - 1, & x \geq 1 \\ 2x^2 - 2, & x < 1\end{matrix}\right. , g(x) = \left\{\begin{matrix}x + 1, & x > 0 \\ -x^2 + 1, & x \leq 0\end{matrix}\right.$$, and $$h(x) = |x|$$, then $$\displaystyle \lim_{x \rightarrow 0} f(g (h (x)))$$ is

A
0

B
1

C
2

D
3

Solution

$$\displaystyle \lim_{x \rightarrow 0^+} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0$$
$$\displaystyle \lim_{x \rightarrow 0^-} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0$$
Therefore, $$\displaystyle \lim_{x \rightarrow 0} f(g(h(x))) = 0$$
Hence, option 'A' is correct.

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 13

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Limits and Continuity Test 13

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SHARING IS CARING

Question 1

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

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

$$2$$

$$\dfrac{1}{3}$$

Solution

Question 2

$$\dfrac {9}{2}$$

$$\dfrac {2}{9}$$

0

$$\dfrac {8}{9}$$

Solution

Question 3

O

$$\infty$$

1

None of these

Solution

Question 4

$$\ln (n!)$$

$$n$$

$$n!$$

$$0$$

Solution

Question 5

16

24

32

8

Solution

Question 6

rational number

irrational number

natural number

complex number

Solution

Question 7

$$2$$

$$-1$$

$$0$$

$$1$$

Solution

Question 8

$$0$$

$$2$$

$$4$$

$$\infty$$

Solution

Question 9

$$0$$

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

$$-1$$

$$1$$

Solution

Question 10

0

1

2

3

Solution

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