Find $\lim_{n \rightarrow \infty} \int_{-1/\sqrt{n}}^0 \sqrt{n} e^{x^2/2} dx= 1 $ Show that $$\lim_{n \rightarrow \infty} \int_{-1/\sqrt{n}}^0 \sqrt{n} e^{x^2/2} dx= 1 $$
The intuition is clear, but I can't think of a way to do this. Thank you.
 A: Since $\;e^{x^2/2}\;$ is continuous everywhere the next function is differentiable:
$$f(x)=\int_0^x\,e^{t^2/2}dt\implies f'(x)=\lim_{x\to 0}\frac{f(x)-f(0)}x=\lim_{x\to 0}\frac1xf(x)\stackrel{\text{since the derivative exists}}=$$
$$=\lim_{n\to\infty}\sqrt n\,f\left(\frac1{\sqrt{n}}\right)=\lim_{n\to\infty}\int_0^{1/\sqrt n}\sqrt n\,e^{t^2/2}\,dt$$
Now just
$$f'(x)=e^{x^2/2}\implies f'(0)=e^0=1$$
Also observe the integrand is an even function.
A: Let $m_n:=e^{1/2n}$. The inequality $1\leq e^{x^2/2}\leq m_n$ for $x\in [-1/\sqrt{n},1]$ implies 
$$
\int_{-1/\sqrt{n}}^0 \sqrt{n} dx\leq I_n:=\int_{-1/\sqrt{n}}^0 \sqrt{n} e^{x^2/2} dx\leq \int_{-1/\sqrt{n}}^0 \sqrt{n}\, m_n dx
$$
or
$1\leq I_n\leq m_n$. The fact that $m_n\to 1$ implies $I_n\to 1$.
A: You can bound the integral and use squeezing:
$$1 = \sqrt{n}e^{0^2/2}\frac1{\sqrt{n}}\le\int_{-1/\sqrt{n}}^0\sqrt{n}e^{x^2/2}\,dx\le\sqrt{n}e^{(-1/\sqrt{n})^2/2}\frac1{\sqrt{n}} = e^{1/2n}\to 1.$$
A: For sure, if you know the antiderivative, the problem becomes simple $$I=\int e^{x^2/2}\,dx=\sqrt{\frac{\pi }{2}} \text{erfi}\left(\frac{x}{\sqrt{2}}\right)$$ where appears the imaginary error function.
So, $$J=\sqrt{n}\int_{-1/\sqrt{n}}^0  e^{x^2/2} dx=\sqrt{\frac{\pi\, n }{2}} \text{erfi}\left(\frac{1}{\sqrt{2n}}\right)$$ The Wikipedia page gives the Maclaurin series $$\text{erfi}(z)=\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)} $$ This being applied to $J$ leads to $$J=1+\frac{1}{6 n}+\frac{1}{40 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
