Tricky function ( show limit is finite) Given the function $f(x,a)=e^{\frac{-x^2}{2a^2}}$, show that $\lim_{a\to 0^+}\frac{f(x,a)}{a}$ is finite?
I was trying with l'Hospital rule but still got an undetermined case. Is there anyone out there who can see some sort of transformation or has an idea for a different approach how to show the limit exist and is finite.
Any references or hints are highly appreciated. Thank you.
 A: Let us set $x=1$ without loss of generality. Substitute $t=\frac{1}{a}$. Then $$\frac{f(x,a)}{a}=t e^{-t^2}=\frac{1}{t^{-1}e^{t^2}}=\frac{1}{\frac{1}{t}+t+2t^3+\cdots}.$$
If $a\rightarrow 0+$, then $t\rightarrow\infty$ and $f(x,a)/a\rightarrow 0$, since the denominator goes to $\infty$.
A: $e^x \ge 1+x$ for any $x > 0$  [which can be proven by comparing derivatives]
For any $a > 0$ and $x \in \mathbb{R}$,
  Let $c=\frac{x^2}{2} > 0$ and so $c/a^2 > 0$
  $0 \le f(x,a) = e^{-c/a^2} = \frac{1}{\large e^{c/a^2}} \le \frac{1}{1+c/a^2}$
  Can you continue from here? (Use the squeeze theorem.)
A: This is a simple case if you invert the variable. Let $a=\frac1b$. Now, you have
$$
\lim_{a\to0^+} \frac{e^{-\frac{x^2}{2a^2}}}a = \lim_{b\to\infty}be^{-x^2b^2/2} = \lim_{b\to\infty}\frac{b}{e^{x^2b^2/2}}\overset{L'hopital}= \lim_{b\to\infty}\frac{1}{x^2be^{x^2b^2/2}} = 0
$$
Note that this applies only for real $x\neq 0$. A simple observation shows that $f(0,a)=1$, and so $\lim_{a\to0^+} \frac1a$ is not finite.
