Is $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$ continuous?. Where is bluff? I have a function defined by $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$. The task is to check, whether $f(x)$ is continuous at $x = 0$. I have proposition of a solution and I would like someone to point out a bluff as there most likely is one - solution is suspisciously too short. 
\begin{equation}
f(x) = \sqrt(x) \sum^{\infty}_{n=1} e^{-n^2x} = \sqrt(x) A_{n}
\end{equation}
I do it because x doesn't change when summing, I treat it as constant. The sum $A_{n}$ is obviously convergent, so there exist some finite M conforming $A_{n}$ < M. Thus, I can write $\lim_{x \rightarrow 0} f(x) < lim_{x \rightarrow 0} xM = 0$
well. that would be finished, but as I have mentioned before, I suspect bluff. 

Note the summation when $x=0$ is infinite since it is an infinite sum of 1's, thus contradicting your argument of finite $M$.
