If f(x)= e^1 x , & if x 0 1, & ifx = 0 find whethe f is continuou… - Vidyadip

Question

If $\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$ find whethe f is continuous at x = 0.

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Answer

Given,
$\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$
We observe
$(\text{LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{e}^\frac{-1}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\bigg(\frac{1}{\text{e}^{\frac{1}{\text{h}}}}\bigg)=\frac{1}{\lim\limits_{\text{h} \rightarrow 0}\text{e}^{\frac{1}{\text{h}}}}=0$
$(\text{RHL at x}=0)=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}(\text{x})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0^+}\text{e}^\frac{1}{\text{h}}=\infty$
Given,
$\text{f}(0)=1$
It is known for a function f(x) to be continuous at x = a,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$
But here,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{0}^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.

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