Limits and Continuity Test 3

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Limits and Continuity Test 3

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statement is not correct

A
$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) } $$

B
$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) } $$

C
$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) } $$

D
$$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } } $$

Solution

All the result other than D are correct.
The last result will hold only when $$g(x)\ne 0$$ and $$\lim\limits_{x\to c} g(x)\ne 0$$ for $$x$$ belonging to a neighbourhood of $$c$$.
A, B, C are general results which are always true.
Question 2
1 / -0

What is the value of $$\underset{x\rightarrow 0}{lim}\dfrac{\sin\,x}{\tan \,3x}$$

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

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

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

D
$$1$$

Solution

Given,

$$\lim_{x\rightarrow 0}\dfrac{\sin \left(x\right)}{\tan \left(3x\right)}$$

applying L'Hospital rule, we get,

$$\lim_{x\rightarrow 0} \dfrac{\cos \left(x\right)}{\sec ^2\left(3x\right)\cdot 3}$$

$$=\dfrac{\cos \left(0\right)}{\sec ^2\left(3\cdot 0\right)\cdot 3}$$

$$=\dfrac{1}{3}$$
Question 3
1 / -0

If $$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$$
Then $$f(2^{+})-f(2^{-})=$$

A
$$0$$

B
$$2$$

C
$$9$$

D
$$4$$

Solution

Given: $$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$$

$$f(2^{+})=\displaystyle \lim _{x\rightarrow 2^+} f(x)$$

$$=\displaystyle \lim _{ h\rightarrow 0}3(2+h)^{2}+4 $$

$$f(2^+)=16$$

$$f(2^{-})=\displaystyle \lim _{x\rightarrow 2^-} f(x)$$

$$=\displaystyle \lim _{ h\rightarrow 0}4(2-h)^{2}-9$$

$$f(2^-)=7$$

So, $$f(2^{+})-f(2^{-})=16-7=9$$
Question 4
1 / -0

If $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\\mathrm{x}+\mathrm{d}(\mathrm{x}\geq\alpha)\end{array}\right.$$is such that
$$\displaystyle \lim_{x\rightarrow \alpha}f(x) =L$$, then $$L=$$

A
$$2d-b$$

B
$$b-d$$

C
$$d+b$$

D
$$b-2d$$

Solution

$$\begin{matrix}lim\\h\rightarrow 0^{-} \end{matrix}\ f(x)=2(\alpha -h)+b=2\alpha +b=L .............(1)\ $$

$$\begin{matrix}lim\\h\rightarrow 0 ^{+} \end{matrix}\ f(x)=(\alpha +h)+d=L\quad \quad \ \alpha =L-d\ ................(2)$$

Substituting value of euation (2)in (1), we get
$$2\left ( L-d \right )+b=L$$
$$L=2d-b$$

Hence, option 'A' is correct.
Question 5
1 / -0

lf $$f(x)=x,x<0;f(x)=0,x=0$$; $$f(x)=x^{2};x>0$$, then $$\displaystyle \lim_{x\rightarrow 0}f(x)$$ is equal to

A
Does not exist

B
$$0$$

C
$$-1$$

D
$$1$$

Solution

Given: $$f(x)$$= $$\left\{ \begin{matrix} x \\ 0 \\ x^ 2 \end{matrix}\begin{matrix}\quad x<0 \\\quad x=0 \\ \quad x>0 \end{matrix} \right\} $$
To Find : $$\underset{x\rightarrow 0}{\lim}$$ $$f(x)$$= ?
Sol: left hand limit $$\rightarrow $$ $$\underset{x\rightarrow 0^-}{\lim}$$ $$f(x)$$=$$\underset{x\rightarrow 0}{\lim}$$ $$x$$ =$$0$$
right hand limit $$\rightarrow $$ $$\underset{x\rightarrow 0^+}{\lim}$$ f(x)=$$\underset{x\rightarrow 0}{\lim}$$ $$x^2$$ =$$0$$,
LHL=RHL
$$f(0)=0$$
$$LHL=RHL=f(0)=0$$
Question 6
1 / -0

$$f(x)=2x+1, a=1,l=3$$ and $$\epsilon=0.001$$, then $$\delta>0$$ satisfying $$0<|x-a|<\delta$$ such that $$|f(x)-l|<\epsilon$$, is

A
$$0.0005$$

B
$$0.005$$

C
$$0.001$$

D
$$0.0001$$

Solution

We have to find $$\delta>0$$, which satisfies $$0<|x-a|<\delta$$ such that $$|f(x)-l|<\epsilon$$
$$\because |f(x)-l| <\epsilon$$
$$\Rightarrow -\epsilon <f(x)-l<\epsilon$$
$$\Rightarrow l-\epsilon<f(x)<l+\epsilon$$
$$\Rightarrow 3-\epsilon<2x+1<3+\epsilon$$
$$\Rightarrow -\epsilon<2x-2<\epsilon$$
$$\Rightarrow -\dfrac{\epsilon }{2}<x-1<\dfrac{\epsilon }{2}$$
$$\Rightarrow |x-1|<\dfrac{\epsilon }{2}$$
$$\Rightarrow |x-a| <\dfrac{\epsilon }{2}$$
$$\delta = \dfrac{\epsilon}{2} = \dfrac{0.001}{2} = 0.0005$$

Hence, option A.
Question 7
1 / -0

Evaluate: $$\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)$$

A
$$\displaystyle \frac{1}{12}$$

B
$$-\displaystyle \frac{4}{3}$$

C
$$-\displaystyle \frac{16}{3}$$

D
$$-\displaystyle \frac{1}{48}$$

Solution

Let $$L =\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right) =\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{\sqrt[3]{8+h}}-\frac{1}{2}\right)$$

Let $$a= \sqrt[3]{8+h}\Rightarrow a^3 = 8+h$$ and $$b=2\Rightarrow b^3 =8$$

Also $$a^3-b^3=(a-b)(a^2+b^2+ab)=8+h-8 =h\Rightarrow $$ (1)

$$\therefore\displaystyle L = \lim_{h\to 0}\frac{1}{h}\left( \frac{1}{a}-\frac{1}{b}\right)=-\lim_{h\to 0}\frac{a-b}{h}\cdot \frac{1}{ab}$$

$$\displaystyle \quad = -\lim_{h\to 0}\frac{a^3-b^3}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$$

$$\displaystyle \quad = -\lim_{h\to 0}\frac{h}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$$ .....using $$(1)$$

As $${h \rightarrow 0}\Rightarrow a^3=8\Rightarrow a=2$$

$$\displaystyle \quad =-\frac{1}{2\times 2(2^2+2^2+2\times 2)}=$$

$$=-\dfrac{1}{48}$$
Question 8
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{(1-e^{x})\sin x}{x^{2}+x^{3}}=$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

The expression can be simplified as,
(lim $$ x \rightarrow 0 \dfrac{(1-{e}^{x})}{x}) \times (\dfrac{sin x}{x})( lim _{ x \rightarrow 0} \dfrac{1}{1+x} )$$
Now using,
lim $$ x \rightarrow 0 \dfrac{({e}^{x} - 1)}{x} = 1 $$
lim $$ x \rightarrow 0 \dfrac{sin x}{x} $$ $$= 1$$
We get,
$$(-1)$$ $$ \times 1 \times lim x \rightarrow 0 \dfrac{1}{1+x} $$
$$= -1$$
Question 9
1 / -0

lf $$\displaystyle \lim_{x\rightarrow a^+}f(x)=L$$, then for each $$\epsilon>0$$, there exists $$\delta>0$$ so that

A
$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon$$

B
$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$$

C
$$a < x < a+\delta\Rightarrow f(x) - L <\epsilon$$

D
$$ a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon$$

Solution

It is fundamental concept that, for limit of a function $$f(x)$$ to exist at any point $$a$$ there exists a real number $$\delta>0$$, such that $$0< |x-a|<\delta$$, for which $$|f(x)-L| < \epsilon$$, where $$\epsilon >0$$
Question 10
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$$

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$\displaystyle \dfrac{1}{2}$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$$

$$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$$

$$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) } $$

$$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$$

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Limits and Continuity Test 3

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Limits and Continuity Test 3

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Question 1

$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) } $$

$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) } $$

$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) } $$

$$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } } $$

Solution

Question 2

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

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

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

$$1$$

Solution

Question 3

$$0$$

$$2$$

$$9$$

$$4$$

Solution

Question 4

$$2d-b$$

$$b-d$$

$$d+b$$

$$b-2d$$

Solution

Question 5

Does not exist

$$0$$

$$-1$$

$$1$$

Solution

Question 6

$$0.0005$$

$$0.005$$

$$0.001$$

$$0.0001$$

Solution

Question 7

$$\displaystyle \frac{1}{12}$$

$$-\displaystyle \frac{4}{3}$$

$$-\displaystyle \frac{16}{3}$$

$$-\displaystyle \frac{1}{48}$$

Solution

Question 8

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 9

$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon$$

$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$$

$$a < x < a+\delta\Rightarrow f(x) - L <\epsilon$$

$$ a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon$$

Solution

Question 10

$$1$$

$$0$$

$$-1$$

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

Solution

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