Paragraph: Let f(x)=\ array ll (e^x-1)/(x), & x > 0 0, & x=0 array...

For right hand continuity at $x=0$\begin{aligned} & \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}\\ &$\begin{array}{l} f(0)=0 \\ \Rightarrow f(x) \text{ is not right continuous at } x=0. \\ \text{ For discontinuity at } x=1 \end{array}$\\ & $\text{ RHL }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \end{aligned}\begin{aligned} & =\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. \\ & =1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots. \\ & \mathrm{LHL}=\lim _{x \rightarrow \mathrm{I}^{-}} 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. \\ & f(1)=\frac{e^1-1}{1}=\left(1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots.\right)-1 \end{aligned}$$=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+$\ldots\Rightarrow $\mathrm{RHL}$ \neq f(1), $\mathrm{LHL}$ \neq f(1)$So$f$is discontinuous.$