Answer
$$ j(x)=e^{\frac{1}{x}} $$
The power of this exponential function is rational function, and rational function is discontinuous wherever the denominator is zero. There is a discontinuity when $x=0$ .
so the function is discontinuous at $a=0$. $$ \lim _{x \rightarrow 0^{+}} j(x)=\infty \quad \quad \text {(limit does not exist) } $$ and $$ \lim _{x \rightarrow 0^{-}} j(x)=0 $$ Since $$ \lim _{x \rightarrow 0^{-}} j(x) \ne \lim _{x \rightarrow 0^{+}} j(x) $$ then, $$ \lim _{x \rightarrow 0} j(x) \quad \quad \text {(does not exist) } $$
Work Step by Step
$$ j(x)=e^{\frac{1}{x}} $$
The power of this exponential function is rational function, and rational function is discontinuous wherever the denominator is zero. There is a discontinuity when $x=0$ .
so the function is discontinuous at $a=0$. $$ \lim _{x \rightarrow 0^{+}} j(x)=\infty \quad \quad \text {(limit does not exist) } $$ and $$ \lim _{x \rightarrow 0^{-}} j(x)=0 $$ Since $$ \lim _{x \rightarrow 0^{-}} j(x) \ne \lim _{x \rightarrow 0^{+}} j(x) $$ then, $$ \lim _{x \rightarrow 0} j(x) \quad \quad \text {(does not exist) } $$
