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NDA Math Limit, Continuity and Differentiability

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Let

f(x)=\begin{cases} \frac{e^x-1}{x} & \text{, } x>0 \\ 0 & \text{, } x=0 \end{cases}

be a real valued function.

[NDA II 2016]

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Question : 63 of 105

Marks: +1, -0

Which of the following statements is/are correct?

f(x)

is right continuous at

x=0

f(x)

is discontinuous at

x=1

Select the correct answer using the code given below:

1 only
2 only
Both 1 and 2
Neither 1 nor 2

Solution:

For right hand continuity at

x=0

RHL=\lim\limits_{x\rightarrow0^{+}}f(x)=\lim\limits_{h\rightarrow0}f(0+h)

=\lim\limits_{h\rightarrow0}\;\frac{e^{h}-1}{h}=\lim\limits_{h\rightarrow0}\;\frac{\left(1+h+\frac{h^{2}}{2!}+\dots\right)-1}{h}

=\lim\limits_{h\rightarrow0}\;\frac{h+\frac{h^{2}}{2!}+\frac{h^{3}}{3!}+\dots}{h}

=\lim\limits_{h\rightarrow0}1+\frac{h}{2!}+\frac{h^{2}}{3!}+\dots=1

f(0)=0

\Rightarrow f(x)

is not right continuous at

x=0

.

For discontinuity at

x=1

RHL=\lim\limits_{x\rightarrow1^{+}}f(x)=\lim\limits_{h\rightarrow0}f(1+h)

=\lim\limits_{h\rightarrow0}\;\frac{e^{1+h}-1}{1+h}

=\lim\limits_{h\rightarrow0}\;\frac{\left(1+(1+h)+\frac{(1+h)^{2}}{2}+\dots\right)-1}{1+h}

=\lim\limits_{h\rightarrow0}\;\frac{(1+h)+\frac{(1+h)^{2}}{2!}+\frac{(1+h)^{3}}{3!}+\dots}{1+h}

=\lim\limits_{h\rightarrow0}1+\frac{1+h}{2!}+\frac{(1+h)^{2}}{3!}+\dots

=1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots

LHL=\lim\limits_{x\rightarrow1^{-}}f(x)=\lim\limits_{h\rightarrow0}f(1-h)

=\lim\limits_{h\rightarrow0}\;\frac{e^{1-h}-1}{1-h}

=\lim\limits_{h\rightarrow0}\;\frac{\left(1+(1-h)+\frac{(1-h)^{2}}{2!}+\dots\right)-1}{1-h}

=\lim\limits_{h\rightarrow0}1+\frac{1-h}{2!}+\frac{(1-h)^{2}}{3!}+\dots

=1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots

f(1)=\frac{e^{1}-1}{1}=\left(1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots\right)-1

=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots

RHL\neq f(1),LHL\neq f(1)

So f is discontinuous.

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