Limits and Continuity Test 29

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Limits and Continuity Test 29

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Weekly Quiz Competition

Question 1
1 / -0

Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}$$.

A
$$\dfrac{1}{\sqrt{2}}$$

B
$$-\dfrac{1}{\sqrt{3}}$$

C
$$-\dfrac{1}{{2}}$$

D
$$-\dfrac{1}{\sqrt{2}}$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}$$

As $$x\rightarrow 0$$, it is $$\dfrac{0}{0}$$ from, so,
using rationalisation

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}\dfrac{(\sqrt{2-x}+\sqrt{2+x})}{(\sqrt{2-x}+\sqrt{2+x})}$$

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x})}}$$

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{-2x}{x(\sqrt{2-x}+\sqrt{2+x})}}=\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{\sqrt2}$$
Question 2
1 / -0

Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 4}\dfrac{2-\sqrt{x}}{4-x}$$.

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

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

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

D
None of these

Solution

$$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{4-\sqrt x}}$$

if $$x\rightarrow 4$$, expression $$\rightarrow \dfrac{0}{0}$$, an indetermined form

using factorisation

$$\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{2^2-(\sqrt x)^2}}=\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{(2-\sqrt x)(2+\sqrt x)}}=\dfrac{1}{4}$$
Question 3
1 / -0

Evaluate the following limits.
If $$\displaystyle\lim_{x\rightarrow a}\dfrac{x^9-a^9}{x-a}=9$$, find all possible values of a.

A
$$2, -2$$.

B
$$1, -1$$.

C
$$1, 0$$.

D
None of these

Solution

Using formula $$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=na^{n-1}$$

$$\displaystyle \lim_{x\rightarrow a}{\dfrac{x^9-a^9}{x-a}}=9.9^8=9$$

$$a^8=1\Rightarrow a=\pm 1$$
Question 4
1 / -0

Evaluate the following limits.
If $$\displaystyle\lim_{x\rightarrow a}\dfrac{x^5-a^5}{x-a}=405$$, find all possible values of a.

A
$$a=3, -3$$

B
$$a=2, -2$$

C
$$a=5, -5$$

D
None of these

Solution

$$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{5}-a^{5}}{x-a}=405$$, find $$a$$ ?
Using $$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1}$$

$$\displaystyle\lim_{x\rightarrow a}\dfrac{x^{5}-a^{5}}{x-a}=5.a^{4}=405$$

$$a^{4}=81$$

$$(a^{2}-9)(a^{2}+9)=0$$

$$a^{2}-9=0$$

$$a=3, -3$$
Question 5
1 / -0

Evaluate the following limits.
$$\displaystyle\lim_{x\rightarrow 2}\dfrac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$$.

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

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

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

D
$$\dfrac{1}{5}$$

Solution

Evaluate $$\displaystyle \lim_{x\to 2}\dfrac {\sqrt {1+4x}-\sqrt {5+2x}}{(x-2)}$$
as $$x\to 2$$, it is $$\dfrac {0}{0}$$ From so,

using rationalisation
$$\displaystyle \lim_{x\to 2}\dfrac {(\sqrt {1+4x}-\sqrt {5+2x}) (1+4x)+\sqrt {5+2x}}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$$

$$\displaystyle \lim_{x\to 2}\dfrac {((1+4x)-(5+2x))}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}$$

$$\displaystyle \lim_{x\to 2}\dfrac {2(x-2)}{(x-2)(6)}=\dfrac {1}{3}$$
Question 6
1 / -0

Evaluate the following limit.
$$\displaystyle\lim_{x\rightarrow 0}\dfrac{8^x-2^x}{x}$$.

A
$$log 4$$

B
$$log 6$$

C
$$log 5$$

D
None of these

Solution

we know $$\displaystyle \lim_{x\rightarrow 0}{\dfrac{a^x-1}{x}}=\log a$$

$$\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-2^x}{x}}=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1+1-2^x}{x}} $$

$$=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1}{x}}-\displaystyle \lim_{x\rightarrow 0}{\dfrac{2^x-1}{x}}=\log 8-\log 2$$

$$=\log 8/2=\log 4$$.
Question 7
1 / -0

If $$f : R \rightarrow (0, \infty)$$ is an increasing function and if $$\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(3x)}{f(x)} = 1$$, then $$\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(2x)}{f(x)}$$ is equal to

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

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

C
$$2$$

D
$$3$$

E
$$1$$

Solution

Given $$f : R \rightarrow (0, \infty)$$ is an increasing function.
And $$\underset{x \rightarrow 2018}{\lim} \dfrac{f(3x)}{f(x)} = 1$$
So, $$\underset{x \rightarrow 2018}{\lim} f(3x) = \underset{x \rightarrow 2018}{\lim} f(x)$$
$$\Rightarrow f(x) = $$ constant.
Therefore,$$ \underset{x \rightarrow 2018}{\lim} \dfrac{f(2x)}{f(x)} = 1$$
Question 8
1 / -0

If $$f$$ is differentiable at $$x = 1$$ and $$\underset{h \rightarrow 0}{\lim} \dfrac{1}{h} f (1 + h) = 5, f'(1) = $$

A
$$0$$

B
$$1$$

C
$$3$$

D
$$4$$

E
$$5$$

Solution

$$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)-f(1)}{h}$$; function is differentiable.

and $$\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$$ [Given function is continuous]
$$\Rightarrow f(1)=0$$

Hence, $$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$$
Question 9
1 / -0

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

A
0

B
$$\infty$$

C
-2

D
2

Solution

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \dfrac{x}{2}} $$
$$ =2 \displaystyle \lim _{x \rightarrow 0}\left[\dfrac{(x / 2)^{2}}{\sin ^{2} \dfrac{x}{2}}\right]\left(\dfrac{e^{x}-1}{x}\right)=2 $$
Question 10
1 / -0

$$ \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}} $$ is equal to

A
$$ -\dfrac{1}{2 \sqrt{2}} $$

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

C
$$ \dfrac{1}{\sqrt{2}} $$

D
Does not exist

Solution

a. $$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}= \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{-x \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}} $$
$$ =-\displaystyle \lim _{x \rightarrow-\infty} \dfrac{\tan \dfrac{1}{x}}{\dfrac{1}{x} \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}=-\dfrac{1}{2 \sqrt{2}} $$

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Limits and Continuity Test 29

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Limits and Continuity Test 29

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SHARING IS CARING

Question 1

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

$$-\dfrac{1}{\sqrt{3}}$$

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

$$-\dfrac{1}{\sqrt{2}}$$

Solution

Question 2

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

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

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

None of these

Solution

Question 3

$$2, -2$$.

$$1, -1$$.

$$1, 0$$.

None of these

Solution

Question 4

$$a=3, -3$$

$$a=2, -2$$

$$a=5, -5$$

None of these

Solution

Question 5

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

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

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

$$\dfrac{1}{5}$$

Solution

Question 6

$$log 4$$

$$log 6$$

$$log 5$$

None of these

Solution

Question 7

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

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

$$2$$

$$3$$

$$1$$

Solution

Question 8

$$0$$

$$1$$

$$3$$

$$4$$

$$5$$

Solution

Question 9

0

$$\infty$$

-2

2

Solution

Question 10

$$ -\dfrac{1}{2 \sqrt{2}} $$

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

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

Does not exist

Solution

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