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Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}, & \mathrm{x}>0 \\ 0, & \mathrm{x}=0\end{array}\right.$ be a real valued function.
Which of the following statements is/are correct?
1- $\mathrm{f}(\mathrm{x})$ is right continuous at $\mathrm{x}=0$.
2- $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1$. Select the correct answer using the code given below:
Options:
Which of the following statements is/are correct?
1- $\mathrm{f}(\mathrm{x})$ is right continuous at $\mathrm{x}=0$.
2- $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1$. Select the correct answer using the code given below:
Solution:
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Verified Answer
The correct answer is:
2 only
For right hand continuity at $x=0$ $\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=\lim _{h \rightarrow 0} \frac{\left(1+h+\frac{h^{2}}{2 !}+\ldots\right)-1}{h} \\=& \lim _{h \rightarrow 0} \frac{h+\frac{h^{2}}{2 !}+\frac{h^{3}}{3 !}+\ldots .}{h} \\=& \lim _{h \rightarrow 0} 1+\frac{h}{2 !}+\frac{h^{2}}{3 !}+\ldots . .=1 \end{aligned}$
$f(0)=0$
$\Rightarrow f(x)$ is not right continuous at $x=0$. For discontinuity at $x=1$
$\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0} \frac{e^{1+h}-1}{1+h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1+h)+\frac{(1+h)^{2}}{2}+\ldots\right)-1}{1+h} \\=& \lim _{h \rightarrow 0} \frac{(1+h)+\frac{(1+h)^{2}}{2 !}+\frac{(1+h)^{3}}{3 !}+\ldots .}{(1+h)} \\=& \lim _{h \rightarrow 0} 1+\frac{(1+h)}{2 !}+\frac{(1+h)^{2}}{3 !}+\ldots . \end{aligned}$
$\begin{aligned}=& 1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \\ \mathrm{LHL} &=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} \frac{e^{(1-h)}-1}{1-h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1-h)+\frac{(1-h)^{2}}{2 !}+\ldots .\right)-1}{(1-h)} \\ &=\lim _{h \rightarrow 0} 1+\frac{(1-h)}{2 !}+\frac{(1-h)^{2}}{3 !}+\ldots . \\ &=1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots . . \end{aligned}$
$f(1)=\frac{e^{1}-1}{1}=\left(1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots .\right)-1$
$\quad=\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots$
$\Rightarrow \mathrm{RHL} \neq f(1), \mathrm{LHL} \neq f(1)$
So $f$ is discontinuous.
$f(0)=0$
$\Rightarrow f(x)$ is not right continuous at $x=0$. For discontinuity at $x=1$
$\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0} \frac{e^{1+h}-1}{1+h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1+h)+\frac{(1+h)^{2}}{2}+\ldots\right)-1}{1+h} \\=& \lim _{h \rightarrow 0} \frac{(1+h)+\frac{(1+h)^{2}}{2 !}+\frac{(1+h)^{3}}{3 !}+\ldots .}{(1+h)} \\=& \lim _{h \rightarrow 0} 1+\frac{(1+h)}{2 !}+\frac{(1+h)^{2}}{3 !}+\ldots . \end{aligned}$
$\begin{aligned}=& 1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \\ \mathrm{LHL} &=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} \frac{e^{(1-h)}-1}{1-h} \\ &=\lim _{h \rightarrow 0} \frac{\left(1+(1-h)+\frac{(1-h)^{2}}{2 !}+\ldots .\right)-1}{(1-h)} \\ &=\lim _{h \rightarrow 0} 1+\frac{(1-h)}{2 !}+\frac{(1-h)^{2}}{3 !}+\ldots . \\ &=1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots . . \end{aligned}$
$f(1)=\frac{e^{1}-1}{1}=\left(1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots .\right)-1$
$\quad=\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots$
$\Rightarrow \mathrm{RHL} \neq f(1), \mathrm{LHL} \neq f(1)$
So $f$ is discontinuous.
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