Question

If \[f\left( x \right) = \begin{cases}e^{1/x} , if & x \neq 0 \\ 1 , if & x = 0\end{cases}\] find whether f is continuous at x = 0.

Answer 1

Given: 

\[f\left( x \right) = \binom{ e^\frac{1}{x} , if x \neq 0}{1, if x = 0}\]

We observe

(LHL at x = 0) = 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]

\[\lim_{h \to 0} e^\frac{- 1}{h} = \lim_{h \to 0} \left( \frac{1}{e^\frac{1}{h}} \right) = \frac{1}{\lim_{h \to 0} e^\frac{1}{h}} = 0\]

(RHL at x = 0) = 

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( h \right)\]

\[\lim_{h \to 0} e^\frac{1}{h} = \infty\]

Given:

\[f\left( 0 \right) = 1\]

It is known that for a function 

\[f\left( x \right)\] to be continuous at x = a, 

\[\lim_{x \to a^-} f\left( x \right) = \lim_{x \to a^+} f\left( x \right) = f\left( a \right)\]

But here,

\[\lim_{x \to 0^-} f\left( x \right) \neq \lim_{x \to 0^+} f\left( x \right)\]

Hence,  

\[f\left( x \right)\]is discontinuous at 

x = 0