I am trying to calculate the following limit:
$$\lim_{t \rightarrow 0} \int_{\frac{x-1}{\sqrt{t}}}^{\frac{x+1}{\sqrt{t}}} exp(\frac{-s²}{4})ds$$ for all $x \in \mathbb R$ where $t \gt0$
For $x=0$, I used the mean-value theorem for integrals and got the following:
$\int_{\frac{-1}{\sqrt{t}}}^{\frac{+1}{\sqrt{t}}} exp(\frac{-s²}{4})ds \le\frac{2}{\sqrt{t}}exp(-\frac{1}{t}) \rightarrow 0$
How do I calculate the limit for any $x \in \mathbb R$ ?
You really should write $\lim_{t\to 0^+}$. since $\sqrt{t}\not\in\mathbb{R}$ if $t<0$. In the $t\to 0^+$ one-sided limit, the integration range becomes $\emptyset$ if $|x|>1$, $\mathbb{R}^+$ if $x=1$, $\mathbb{R}$ if $x\in (-1,\,1)$, or $\mathbb{R}^-$ if $x=-1$. This multiplies $\int_\mathbb{R}\exp\frac{-s^2}{4}ds=2\sqrt{\pi}$ by $0,\,\frac{1}{2},\,1,\,\frac{1}{2}$ depending on the case considered. Note in particular that if $|x|>1$ the integral is of width $\frac{2}{\sqrt{t}}$, but with an integrand of modulus $\le\exp -\frac{(|x|-1)^2}{4}$, which is so small the integral $\to 0$.