Continuity of a piecewise function with an essential singularity

For $x=a+h$ with $h\neq0$,
$$f(a+h)=h\cdot\frac{e^{1/h}}{1+e^{1/h}}.$$
Since $e^{1/h}>0$, we have $0<\dfrac{e^{1/h}}{1+e^{1/h}}<1$. Hence
$$|f(a+h)-f(a)|=|f(a+h)|<|h|.$$
As $h\to0$, $|h|\to0$, so $f(a+h)\to f(a)=0$. Therefore $f$ is continuous at $a$.

Conclusion. The two one-sided limits both equal $0$, which equals $f(a)$. Hence $f$ is continuous at $x=a$.