$ \lim_{t\to 0} \int_{|x|>\epsilon} \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2 \pi t}} dx=0 $ I want to prove that: \begin{equation}
  \lim_{t\to 0} \int_{|x|>\epsilon} \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2 \pi t}} dx=0,
\end{equation}
for any $\epsilon >0$
I've shown using polar coordinates that \begin{equation} \int \limits_{-\infty}^{+\infty} \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2 \pi t}}=1.
\end{equation}
I tried the same approach for the limit I need to prove. But I obtained a result that depended on $\epsilon$ only, and not on $t$, and therefore it does not go to $0$. Where is my mistake?
Other ways to approach it would be greatly appreciated too (preferably, with real analysis methods only).
 A: We can use the same polar coordinates trick here and in fact obtain a good bound on the integral. Let
$$I(\epsilon,t) = \int_{\vert x \vert > \epsilon} \dfrac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx$$
We now have
$$I^2(\epsilon,t) = \int_{\vert x \vert > \epsilon} \dfrac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx \cdot \int_{\vert y \vert > \epsilon} \dfrac{e^{-\frac{y^2}{2t}}}{\sqrt{2\pi t}}dy = \int_{\vert x \vert > \epsilon} \int_{\vert y \vert > \epsilon} \dfrac{e^{-\frac{x^2+y^2}{2t}}}{2\pi t}dxdy \leq \int_{x^2+y^2 > \epsilon^2} \dfrac{e^{-\frac{x^2+y^2}{2t}}}{2\pi t}dxdy$$
Now moving to polar coordinates, we have
$$I^2(\epsilon,t) \leq \int_{\theta=0}^{2\pi} \int_{r=\epsilon}^{\infty} \dfrac{e^{-\frac{r^2}{2t}}}{2\pi t}rdrd\theta = \dfrac1t \int_{r=\epsilon}^{\infty}re^{-\frac{r^2}{2t}}dr = \dfrac1t \cdot te^{-\frac{\epsilon^2}{2t}} = e^{-\frac{\epsilon^2}{2t}}$$
This gives us that
$$\boxed{\color{blue}{I(\epsilon,t) \leq e^{-\frac{\epsilon^2}{4t}}}}$$
Take $\lim t \to 0$ to obtain what you want.
