Limits and Continuity Test 15

Result

Limits and Continuity Test 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$|x| < 1$$, then $$\displaystyle \lim_{n \rightarrow \infty }\{ (1 + x) (1+x^2)(1 + x^4) ..... (1 + x^{2n}) \}$$ is equal to

A
$$\displaystyle \frac{1}{x - 1}$$

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

C
$$1- x$$

D
$$x -1$$

Solution

We have,
$$\displaystyle \lim_{n \rightarrow \infty} \{ (1 + x)(1 + x^2)(1 + x^4) .... (1 + ^{2n})\}$$
$$=\displaystyle \lim_{n \rightarrow \infty} \left \{ \frac{(1 - x)(1 +x)(1 + x^2) .... (1 + x^{2n})}{1 -x} \right\}$$

$$= \displaystyle \lim_{n \rightarrow \infty} \frac{1 - x^{4n}}{1 - x}$$

$$= \displaystyle \frac{1 - 0}{1 - x} = \frac{1}{1- x}$$ $$\left [ \because \lim_{n \rightarrow \infty} x^{4n} = 0 for -1 < x < 1 \right ]$$
Hence, option 'B' is correct.
Question 2
1 / -0

The value of $$\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$ for $$(a>1)$$ is equal to?

A
$$1$$

B
$$0$$

C
$$\displaystyle\frac{\pi}{2}$$

D
Does not exist

Solution

$$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$

$$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$$
Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.
Question 3
1 / -0

The value of
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 \sin^2 x + \sin x-1}{2 \sin^2 x - 3 \sin x + 1} $$

A
$$3$$

B
$$-3$$

C
$$6$$

D
$$0$$

Solution

We have
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 sin^2 x + sin x -1}{2 sin^2 x-3 sin x + 1}$$

$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{(2 sin x - 1)(sin x + 1)}{(2 sin x - 1)(sin x - 1)}$$

$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{sin x + 1}{sin x - 1} = - 3$$
Question 4
1 / -0

Let $$f(x)=\sin x$$, $$g(x)=\left [ x+1 \right ]$$ and $$g(f(x))=h(x)$$, where [.] is the greatest integer function. Then $$h^+\left ( \displaystyle \dfrac{\pi }{2} \right )$$ is

A
non existent

B
$$1$$

C
$$-1$$

D
none of these

Solution

$$f(x) =\sin(x)$$ and $$g(x) = [1+x]$$

$$\Rightarrow g(f(x)) = [1+\sin x]=h(x)$$

$$h^{+}\left( \cfrac{\pi}{2}\right) =[1+\sin(\pi/2)]= 1$$
Question 5
1 / -0

Which one of the following statement is true?

A
if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

B
if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exist

C
if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

D
if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exists
Solution
The statements can be proved false by giving counter examples

(1) If $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists
Let $$g(x) = 1/x $$ and $$f(x) = 2x$$, with $$c = 0$$,
then $$\displaystyle \lim_{x\to c}f(x).g(x) =2 $$
$$\displaystyle \lim_{x\to c}f(x) = 0 $$ ,but $$\displaystyle \lim_{x\to c}g(x)$$ does not exist
(2) Same example as (1)
(4) Let $$g(x) =\dfrac{ x+ 1}{x} $$ and $$f(x) = \dfrac{x-1}{x} $$, with $$c = 0$$,
then $$\displaystyle \lim_{x\to c}[f(x)+g(x)] =2$$
but $$\displaystyle \lim_{x\to c}f(x) \ and \ g(x)$$ do not exist
Question 6
1 / -0

$$\displaystyle \lim_{n\to\infty }\frac{n^{p}\sin ^{2}\left ( n! \right )}{n+1}$$, $$0<p<1$$, is equal to

A
$$0$$

B
$$\infty $$

C
$$1$$

D
$$none\ of\ these$$

Solution

We have,
$$\displaystyle \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n + 1} = \lim_{n \rightarrow \infty} \dfrac{sin^2 (n!)}{1 + \dfrac{1}{n}} \times n^{p - 1}$$
$$\Rightarrow \displaystyle \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n + 1} = \lim_{n \rightarrow \infty} \dfrac{sin^2 (n!)}{n^{1 - p}} \times \dfrac{1}{1 + \dfrac{1}{n}}$$
$$\displaystyle \Rightarrow \lim_{n \rightarrow \infty} \dfrac{n^p sin^2 (n!)}{n+1} = \dfrac{\text{An oscillating number}}{\infty} \times 1 = 0 \times 1 = 0$$.
Hence, option 'A' is correct.
Question 7
1 / -0

$$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$$ and $$g\left( x \right) =\begin{cases} { x }^{ 2 }+1\qquad ;\qquad x\neq 0 \\ 4\qquad \qquad ;\qquad x=0 \end{cases}.$$
Then $$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right)\right)} $$ is

A
$$1$$

B
$$4$$

C
$$5$$

D
non-existent

Solution

$$\lim _{ x\rightarrow 0 }{ g\left( f\left( x \right) \right) } =g\left( f\left( 0 \right) \right) $$
$$=g(2)={ 2 }^{ 2 }+1=5$$
Question 8
1 / -0

Which one of the following statements is true?

A
If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exists.

B
If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exist.

C
If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exists.

D
If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exist.

Solution

For option A and B, take $$f(x)=x$$ and $$g(x)=\dfrac { 1 }{ x }$$.
Limit of $$g(x)$$ doesn't exist at $$x=0$$, but for $$\displaystyle \lim _{ x\to 0} f(x).g(x)$$ and $$f(x)=x$$ limit exists.

For option C,
Lets assume $$h(x) = f(x)+g(x)$$,
Take limit on both sides as $$x\to c$$
$$\lim_{x\to c}h(x) = \lim_{x\to c}(f(x)+g(x))$$
And it is given that $$\lim_{x\to c}f(x)$$ exists.
Using sum law of limits:
If $$\lim_{x\to c}F(x)$$ and $$\lim_{x\to c}G(x)$$ exists, then $$\lim_{x\to c}(F(x)\pm G(x))$$ also exists.

$$\therefore \lim_{x\to c}(h(x) - f(x)) = \lim_{x\to c}(f(x) + g(x) - f(x)) = \lim_{x\to c}g(x)$$ exists.

For option D, take $$f(x)=-\dfrac{cosx}{x^2}$$ and $$g(x)=\dfrac { 1 }{ x^2 }$$.
Limit of $$g(x)$$ and $$f(x)=-\dfrac{cosx}{x^2}$$ doesn't exist at $$x=0$$, but for $$\displaystyle \lim _{ x\to 0} f(x)+g(x)$$ limit exists.

Hence option C.
Question 9
1 / -0

If $$\displaystyle \lim_{x\rightarrow 0}(f(x)\:g(x))$$ exists for any functions $$f$$ and $$g$$ then

A
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

B
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

C
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ may not exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

D
$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

Solution

Take $$\displaystyle f(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$$ and $$\displaystyle g(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$$. Then $$f(x)\:g(x)=1$$ for $$x\neq 0$$. Hence $$\displaystyle \displaystyle \lim_{x\rightarrow 0}f(x)$$ and $$\displaystyle \lim_{x\rightarrow 0}g(x)$$ does not exist but $$\displaystyle\lim_{x\rightarrow 0}f(x)\:g(x)=1$$
Hence, option 'D' is correct.
Question 10
1 / -0

If $$p\left( x \right) ={ a }_{ 0 }+{ a }_{ 1 }x+...+{ a }_{ n }{ x }^{ n }$$ and $$\left| p\left( x \right) \right| \le \left| { e }^{ x-1 }-1 \right| $$ for all $$x\ge 0,$$ then $$\left| { a }_{ 1 }+2{ a }_{ 2 }+3{ a }_{ 3 }+...+n{ a }_{ n } \right| $$

A
$$\le 1$$

B
$$\ge 1$$

C
$$\ge 0$$

D
$$\le 0$$

Solution

We have, $$\displaystyle p'\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \frac { p\left( 1+h \right) -p\left( 1 \right) }{ h } } \Rightarrow \left| p'\left( 1 \right) \right| =\left| \lim _{ h\rightarrow 0 }{ \frac { p\left( 1+h \right) -p\left( 1 \right) }{ h } } \right| $$

$$\displaystyle =\lim _{ h\rightarrow 0 }{ \frac { \left| p\left( 1+h \right) -p\left( 1 \right) \right| }{ \left| h \right| } } \le \lim _{ h\rightarrow 0 }{ \frac { \left| p\left( 1+h \right) +p\left( 1 \right) \right| }{ \left| h \right| } } $$ ........(1)

Now, $$\left| p\left( x \right) \right| \le \left| { e }^{ x-1 }-1 \right| \Rightarrow \left| p\left( 1-h \right) \right| \le \left| { e }^{ h }-1 \right| $$ and $$\left| p\left( 1 \right) \right| \le 0\Rightarrow p\left( 1 \right) =0$$

Hence, $$\displaystyle \left| p'\left( 1 \right) \right| \le \lim _{ h\rightarrow 0 }{ \frac { \left| { e }^{ h }-1 \right| }{ \left| h \right| } } =1$$

$$\displaystyle \Rightarrow \left| p'\left( 1 \right) \right| \le 1\Rightarrow \left| { a }_{ 1 }+2{ a }_{ 2 }+3{ a }_{ 3 }+...+n{ a }_{ n } \right| \le 1$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 15

Result

Limits and Continuity Test 15

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

$$1- x$$

$$x -1$$

Solution

Question 2

$$1$$

$$0$$

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

Does not exist

Solution

Question 3

$$3$$

$$-3$$

$$6$$

$$0$$

Solution

Question 4

non existent

$$1$$

$$-1$$

none of these

Solution

Question 5

if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exist

if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exists

Solution

Question 6

$$0$$

$$\infty $$

$$1$$

$$none\ of\ these$$

Solution

Question 7

$$1$$

$$4$$

$$5$$

non-existent

Solution

Question 8

If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exists.

If $$\displaystyle\lim_{x\rightarrow c}{f(x).g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ exist.

If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ exist, then $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exists.

If $$\displaystyle\lim_{x\rightarrow c}{f(x)+g(x)}$$ exists, then $$\displaystyle\lim_{x\rightarrow c}{f(x)}$$ and $$\displaystyle\lim_{x\rightarrow c}{g(x)}$$ also exist.

Solution

Question 9

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ may not exist but $$\displaystyle \lim_{x\rightarrow a}g(x)$$ exist

$$\displaystyle \lim_{x\rightarrow a}f(x)$$ and $$\displaystyle \lim_{x\rightarrow a}g(x)$$ may not exist

Solution

Question 10

$$\le 1$$

$$\ge 1$$

$$\ge 0$$

$$\le 0$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.