Limits and Continuity Test 35

Result

Limits and Continuity Test 35

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Question 1
1 / -0

Consider $$f(x)=\lim _{ n\rightarrow \infty }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } } }{ { x }^{ n }+\sin { { x }^{ n } } } } $$ for $$x>0,x\neq 1,f(1)=0$$ then

A
$$f$$ is continuous at $$x=1$$

B
$$f$$ has a discontinuity at $$x=1$$

C
$$f$$ has an infinite or oscillatory discontinuity at $$x=1$$

D
$$f$$ has a removal type of discontinuity at $$x=1$$
Question 2
1 / -0

If $$\sum _{ r=1 }^{ k }{ \cos ^{ -1 }{ \beta } } =\cfrac { k\pi }{ 2 } $$ for any $$k\ge 1$$ and $$A=\sum _{ r=1 }^{ k }{ { \left( { \beta }_{ r } \right) }^{ r } } $$, then $$\lim _{ x\leftarrow A }{ \cfrac { { \left( 1+x \right) }^{ 1/3 }-{ \left( 1-2x \right) }^{ 1/4 } }{ x+{ x }^{ 2 } } } $$ is equal to

A
$$0$$

B
$$\cfrac{1}{2}$$

C
$$\cfrac { \pi }{ 2 } $$

D
$$\cfrac { 5 }{ 6 } $$

Solution
Question 3
1 / -0

$$\displaystyle \lim_{x \rightarrow 0} \frac{ae^x + b cos x + c. e^{-x}}{sin^2 x} = 4$$ then b =

A
2

B
4

C
-2

D
-4
Question 4
1 / -0

$$\lim_{x\rightarrow \dfrac{\pi}{6}} \dfrac{2 sin^{2} x + sin x - 1}{2 sin^{2} x - 3 sin x + 1}$$

A
6

B
-6

C
-3

D
3

Solution
Question 5
1 / -0

$$\mathop {\lim }\limits_{x \to {a^ + }} {{\left\{ x \right\}\sin \left( {x - a} \right)} \over {{{\left( {x - a} \right)}^2}}}$$
is equal to (where {.} denotes the fraction
part of x and $$a \in N$$

A
0

B
1

C
does not exist

D
none of thes

Solution
Question 6
1 / -0

Let$$f(\theta) = \dfrac{1}{tan^{9}\theta} {(1+tan\theta)^{10}+(2+tan\theta)^{10}+....+(20+tan\theta)^{10}}-20tan\theta$$. The left hand limit of $$f(\theta)$$ as $$\theta \rightarrow \dfrac{\pi}{2}$$ is:

A
1900

B
2000

C
2100

D
2200

Solution
Question 7
1 / -0

If $$l=\lim\limits_{n\to 3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$$ and $$m=\lim\limits_{n\to -3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$$, then

A
$$l\ne m$$

B
$$l=2m$$

C
$$l=-m$$

D
$$l=m$$

Solution

Given $$l=\underset{x\to3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore, $$l=\underset {x\to3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to 3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$$

Given $$m=\underset{x\to-3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore, $$m=\underset {x\to-3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to -3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to-3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$$

Hence, $$l=8,m=8$$

Therefore, $$l= m$$
Question 8
1 / -0

If $$f : R \rightarrow R$$ is defined by
$$f(x) = \left \{\begin{matrix} \dfrac{x + 2}{x^2 + 3x + 2} & if & x \in R - \{-1, -1\} \\ -1 & if & x = -2 \\ 0 & if &x = -1\end{matrix} \right.$$ then $$f(x)$$ continuous on the set

A
$$R$$

B
$$R - \{-2\}$$

C
$$R - \{-1, -2\}$$

D
$$R-\{-1\}$$

Solution
Question 9
1 / -0

If $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{1 - \left| x \right|}}{{1 + x}},{\rm{ }}x \ne - 1\\1,{\rm{ }}x = - 1{\rm{ }}\end{array} \right.$$ then $$f\left( {\left[ {2x} \right]} \right),$$ where $$\left[ {} \right]$$ represents the greatest integer function , is

A
discontinuous at $$x = - 1$$

B
continuous at $$x = 0$$

C
continuous at $$x = \frac{1}{2}$$

D
continuous at $$x = 1$$

Solution
Question 10
1 / -0

if $$f\left( x \right) = \left\{ {\matrix{ {\cos \left[ x \right],} & {x \ge 0} \cr {\left| x \right| + a,} & {x < 0} \cr
} } \right\}$$ Find
the value of a , given that $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ exists,
where[.] denotes

A
-1

B
2

C
1

D
0

Solution