Rd sharma solution class 12 chapter Continuity

Exercise 9.1

Question 1

Test the continuity of the function on f(x) at the origin:

f(x)=\begin{cases}\dfrac{x}{|x|},x\not=0\\~~1~,x=0\end{cases}

Sol :

Given

f(x)=\begin{cases}\dfrac{x}{|x|},x\not=0\\~~1~,x=0\end{cases}

We observe

(LHL at x= 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~f(-h)

=\lim_{h\to0}~\dfrac{-h}{\lvert -h \rvert}

=\lim_{h\to0}~\dfrac{-h}{h}

=\lim_{h\to0}~-1

=-1

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~f(h)

=\lim_{h\to0}~\dfrac{h}{\lvert h \rvert}

=\lim_{h\to0}~\dfrac{h}{h}

=\lim_{h\to0}~1

=1

\lim_{x\to0^{+}}~f(x)\not=\lim_{x\to0^{-}}~f(x)

Hence, f(x) is discontinuous at the origin.

 

Question 2

A function f(x) is defined as 

f(x)=\left\{\begin{matrix}{\dfrac{x^2-x-6}{x-3}},x\not=3\\~~~~~~~~~~5~~~~~~,x=3\end{matrix}\right.

Show that f(x) is continuous that x=3

Sol :

Given

f(x)=\left\{\begin{matrix}{\dfrac{x^2-x-6}{x-3}},x\not=3\\~~~~~~~~~~5~~~~~~,x=3\end{matrix}\right.

We observe

(LHL at x = 3)

=\lim_{x\to3^{-}}~f(x)

=\lim_{h\to0}~f(3-h)

=\lim_{h\to0}~\dfrac{(3-h)^2-(3-h)-6}{(3-h)-3}

=\lim_{h\to0}~\dfrac{9+h^2-6h-3+h-6}{-h}

=\lim_{h\to0}~\dfrac{h^2-5h}{-h}

=\lim_{h\to0}~(5-h)

= 5

 

(RHL at x= 3)

=\lim_{x\to3^{+}}~f(x)

=\lim_{h\to0}~f(3+h)

=\lim_{h\to0}~\dfrac{(3+h)^2-(3+h)-6}{(3+h)-3}

=\lim_{h\to0}~\dfrac{9+h^2+6h-3-h-6}{h}

=\lim_{h\to0}~\dfrac{h^2+5h}{h}

=\lim_{h\to0}~(5+h)

= 5

Also , f (3) = 5

\lim_{x\to3^{+}}~f(x)=\lim_{x\to3^{-}}~f(x)=f(3)

Hence, f(x) is continuous at x=3

 

Question 3

A function f(x) is defined as 

f(x)=\left\{\begin{matrix}{\dfrac{x^2-9}{x-3}},x\not=3\\~~~~~~~6~~~,x=3\end{matrix}\right.

Show that f(x) is continuous at x = 3

Sol :

Given 

f(x)=\left\{\begin{matrix}{\dfrac{x^2-9}{x-3}},x\not=3\\~~~~~~~6~~~,x=3\end{matrix}\right.

We observe

(LHL at x = 3)

=\lim_{x\to3^{-}}~f(x)

=\lim_{h\to0}~f(3-h)

=\lim_{h\to0}~\dfrac{(3-h)^2-9}{{3-h}-3}

=\lim_{h\to0}~\dfrac{3^2+h^2-6h-9}{(3-h)-3}

=\lim_{h\to0}~\dfrac{h^2-6h}{-h}

=\lim_{h\to0}~\dfrac{h(h-6)}{-h}

=\lim_{h\to0}~(6-h)

= 6

 

(RHL at x = 3)

=\lim_{x\to3^{+}}~f(x)

=\lim_{h\to0}~f(3+h)

=\lim_{h\to0}~\dfrac{(3+h)^2-9}{{3+h}-3}

=\lim_{h\to0}~\dfrac{h^2+6h}{h}

=\lim_{h\to0}~\dfrac{h(h+6)}{h}

=\lim_{h\to0}~(6+h)

= 6

Also , f (3) = 6

\lim_{x\to3^{-}}~f(x)=\lim_{x\to3^{+}}~f(x)=f(3)

Hence, f(x) is continuous at x=3

 

Question 4

If f(x)=\left\{\begin{matrix}{\dfrac{x^2-1}{x-1}},x\not=1\\~~~~2~~~,x=1\end{matrix}\right.

Find whether f(x) is continuous at x=1

Sol :

Given

f(x)=\left\{\begin{matrix}{\dfrac{x^2-1}{x-1}},x\not=1\\~~~~2~~~,x=1\end{matrix}\right.

We observe

(LHL at x = 1)

=\lim_{x\to1^{-}}~f(x)

=\lim_{h\to0}~f(1-h)

=\lim_{h\to0}~\dfrac{(1-h)^2-1}{(1-h)-1}

=\lim_{h\to0}~\dfrac{1+h^2-2h-1}{(1-h)-1}

=\lim_{h\to0}~\dfrac{h^2-2h}{-h}

=\lim_{h\to0}~\dfrac{h(h-2)}{-h}

=\lim_{h\to0}~(2-h)

= 2

 

(RHL at x = 1)

=\lim_{x\to1^{+}}~f(x)

=\lim_{h\to0}~f(1+h)

=\lim_{h\to0}~\dfrac{(1+h)^2-1}{{1+h}-1}

=\lim_{h\to0}~\dfrac{1+h^2+2h-1}{(1+h)-1}

=\lim_{h\to0}~\dfrac{h^2+2h}{h}

=\lim_{h\to0}~\dfrac{h(h+2)}{h}

=\lim_{h\to0}~(2+h)

= 2

 

Also , f (1) = 2

\lim_{x\to1^{-}}~f(x)=\lim_{x\to1^{+}}~f(x)=f(1)

Hence, f(x) is continuous at x=1

 

Question 5

If f(x)=\left\{\begin{matrix}{\dfrac{sin~3x}{x}},x\not=0\\~~~~1~~~~~,x=0\end{matrix}\right.

Find whether f(x) is continuous at x=0

Sol :

Given

If f(x)=\left\{\begin{matrix}{\dfrac{sin~3x}{x}},x\not=1\\~~~~1~~~~~,x=1\end{matrix}\right.

We observe

(LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~f(-h)

=\lim_{h\to0}~\dfrac{sin~(-3h)}{-h}

=\lim_{h\to0}~-\dfrac{sin~(3h)}{-h}

=\lim_{h\to0}~\dfrac{3~sin~(3h)}{3h}

=3~\lim_{h\to0}~\dfrac{sin~(3h)}{3h}

= 3\times1

= 3

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~f(h)

=\lim_{h\to0}~\dfrac{sin~(3h)}{h}

=\lim_{h\to0}~\dfrac{3~sin~(3h)}{3h}

=3~\lim_{h\to0}~\dfrac{sin~(3h)}{3h}

= 3\times1

= 3

 

Also , f (0) = 1

\lim_{x\to0^{-}}~f(x)=\lim_{x\to0^{+}}~f(x)\not=f(0)

Hence, f(x) is discontinuous at x=0

 

 

Question 6

If f(x)=\Bigg\{\begin{matrix}e^{1/x},x\not=0\\2~~~~,x=0\end{matrix}

Find whether f(x) is continuous at x=0

Sol :

Given

If f(x)=\Bigg\{\begin{matrix}e^{1/x},x\not=0\\2~~~~,x=0\end{matrix}

We observe

(LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~f(-h)

=\lim_{h\to0}~e^\frac{-1}{~h}

=\lim_{h\to0}~\Bigg(\dfrac{1}{e^\frac{1}{~h}}\Bigg)

=\dfrac{1}{e^{\infty}}

=\dfrac{1}{\infty}

= 0

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~f(h)

=\lim_{h\to0}~e^\frac{1}{~h}

=\lim_{h\to0}~e^\infty

= 0

 

Also , f (0) = 1

\lim_{x\to0^{-}}~f(x)=\lim_{x\to0^{+}}~f(x)\not=f(1)

Hence, f(x) is discontinuous at x=0

 

 

Question 7

Let f(x)=\left\{\begin{matrix}{\dfrac{1-cos~x}{x^2}},x\not=0\\~~~~1~~~~~~~~~~,x=0\end{matrix}\right.

Show that f(x) is discontinuous at x=0

Sol :

Given

Let f(x)=\left\{\begin{matrix}{\dfrac{1-cos~x}{x^2}},x\not=0\\~~~~1~~~~~~~~~~,x=0\end{matrix}\right.

Consider

(LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~\dfrac{1-cos~(-h)}{(-h)^2}   [\because cos(-\theta)=cos~\theta]

=\lim_{h\to0}~\dfrac{1-cos~h}{h^2}   [\because 1-cos~\theta=2sin^2\dfrac{\theta}{2}]

=\lim_{h\to0}~\dfrac{2sin^2~\dfrac{h}{2}}{h^2}   

=\lim_{h\to0}~2\Bigg({\dfrac{sin~\dfrac{h}{2}}{h}}\Bigg)^2   

=\lim_{h\to0}~2\Bigg({\dfrac{sin~\dfrac{h}{2}}{2\times\dfrac{h}{2}}}\Bigg)^2   

=\dfrac{1}{2}

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~\dfrac{1-cos~h}{h^2}   [\because cos(-\theta)=cos~\theta]

=\lim_{h\to0}~\dfrac{1-cos~h}{h^2}   [\because 1-cos~\theta=2sin^2\dfrac{\theta}{2}]

=\lim_{h\to0}~\dfrac{2sin^2~\dfrac{h}{2}}{h^2}   

=\lim_{h\to0}~2\Bigg({\dfrac{sin~\dfrac{h}{2}}{h}}\Bigg)^2   

=\lim_{h\to0}~2\Bigg({\dfrac{sin~\dfrac{h}{2}}{2\times\dfrac{h}{2}}}\Bigg)^2   

=\dfrac{1}{2}

Also , f (0) = 1

\lim_{x\to^0{-}}~f(x)=\lim_{x\to0^{+}}~f(x)\not=f(1)

Hence, f(x) is discontinuous at x=0

 

 

Question 8

Show that f(x)=\left\{\begin{matrix}{\dfrac{x-|x|}{2}},x\not=0\\~~~~2~~~~,x=0\end{matrix}\right.

is continuous at x=0

Sol :

Given

If f(x)=\left\{\begin{matrix}{\dfrac{x-|x|}{2}},x\not=0\\~~~~2~~~~,x=0\end{matrix}\right.

We observe

(LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~\dfrac{-h-|-h|}{2}

=\lim_{h\to0}~\dfrac{-h-h}{2}

= 0

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~\dfrac{h-|h|}{2}

= 0

Also , f (0) = 2

\lim_{x\to0^{-}}~f(x)=\lim_{x\to0^{+}}~f(x)\not=f(1)

Hence, f(x) is continuous at x=0

 

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