Rigorous proof of a limit Guys I'm feeling a bit rusty in analysis... How do I prove that
$$\lim_{y \to 0^+} \frac{e^{- x^2/y}}{\sqrt y} = 0$$
With $x\neq 0$. I have tried the approach with $\epsilon$ and $\delta$ but I've got nothing out of it... Help would be greatly appreciated!
 A: Enforcing the substitution $y \to 1/t^2$ reveals
$$\lim_{y\to 0^+}\frac{e^{-x^2/y}}{\sqrt{y}}=\lim_{t\to\infty}te^{-x^2t^2}=0$$
for $x\ne0$ since for any $\epsilon>0$,
$$\begin{align}
\left|te^{-x^2t^2}\right|&\le \frac{|t|}{1+x^2t^2}\\\\
&<\frac{|t|}{t^2x^2}\\\\
&=\frac{1}{x^2|t|}\\\\
&<\epsilon
\end{align}$$
whenever $|t|>B=\frac{1}{x^2\epsilon}$ or equivalently whenever $0<y<\delta=|x|\sqrt{\epsilon}$
A: HINT:
From $e^x = x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots$, you have:
$$ e^{-\dfrac{x^2}{y}} = \dfrac{1}{\frac{x^2}{y} + \frac{\left(\frac{x^2}{y}\right)^2}{2} + \frac{\left(\frac{x^2}{y}\right)^3}{3!} \cdots} \\
 = \dfrac{y}{x^2 + \frac{x^4}{2y} + \cdots}$$
A: ale42 had the right answer, intuitively-speaking, in the comments. I slightly rigorous justification is that (for a couple steps below, we will need that $y > 0$)
$$
\lim_{y \to 0^+} \left| \frac{e^{- x^2/y}}{\sqrt y} \right|
= 
\lim_{y \to 0^+} \frac{1}{\sqrt{y} \left| (1 + \frac{x^2}{y} + \frac{x^4 / y^2}{2!} + \cdots) \right| }
=
\lim_{y \to 0^+} \frac{1}{\sqrt{y} (1 + \left| \frac{x^2}{y} \right| + \left| \frac{x^4 / y^2}{2!} \right| + \cdots) } 
\leq
\lim_{y \to 0^+} \frac{1}{\sqrt{y} (1 + \left| \frac{x^2}{y} \right|) }
=
\lim_{y \to 0^+} \frac{\sqrt{y}}{y + x^2 }
\to 0 \text{ if } x \neq 0.
$$
and so 
$$
\lim_{y \to 0^+} \frac{e^{- x^2/y}}{\sqrt y} = 0
$$
A: $$\lim_{y \to 0^+} \frac{e^{- x^2/y}}{\sqrt y}=\lim_{t \to \infty } \frac{\sqrt t}{e^{tx^2}} = \lim_{t \to \infty }\frac{1}{\sqrt t} \frac{t}{e^{tx^2}} \tag1$$
Using L'Hospital:
$$
\lim_{t \to \infty }\frac{t}{e^{tx^2}} = \lim_{t \to \infty }\frac{1}{x^2e^{tx^2}}=0 
$$
and : $$\lim_{t \to \infty }\frac{1}{\sqrt t}=0$$
from (1) we get the result.
