Prove that $\lim_{n\to\infty}\int_\delta^\infty ne^{-n^2x^2}\text{d}x = 0$ for each positive $\delta$. I'm having trouble with showing that $\lim_{n\to\infty}\int_\delta^\infty ne^{-n^2x^2}\text{d}x = 0$. How would one prove this? I see that it is equivalent to proving that $\lim_{n\to\infty}\int_{-\delta}^\delta ne^{-n^2x^2}\text{d}x = 1$ because of the fact that $\int_{-\infty}^\infty ne^{-n^2x^2}\text{d}x = \sqrt{\pi}$. I also see that for every nonzero $x$ the integrand goes to zero very rapidly, but I really have no clue how to find a proof for the limit(s) above.
Help is appreciated!
 A: $e^{u}\geq u$ for $u>0$, so $ne^{-(nx)^{2}}\leq\dfrac{n}{n^{2}x^{2}}=\dfrac{1}{nx^{2}}\leq\dfrac{1}{x^{2}}$ for all $x\geq\delta$, now $x^{-2}\in L^{1}(\delta,\infty)$ so Lebesgue Dominated Convergence Theorem applies.
Also note that $\lim_{n\rightarrow\infty}ne^{-n^{2}x^{2}}=0$.
Perhaps, change of variable gives $\displaystyle\int_{\delta}^{\infty}ne^{-n^{2}x^{2}}dx=\int_{n\delta}^{\infty}e^{-x^{2}}dx=\int_{0}^{\infty}e^{-x^{2}}dx-\int_{0}^{n\delta}e^{-x^{2}}dx$. Now Monotone Convergence Theorem gives $\displaystyle\int_{0}^{n\delta}e^{-x^{2}}dx\rightarrow\int_{0}^{\infty}e^{-x^{2}}dx$.
A: $$I=\int_\delta^\infty ne^{-n^2x^2}\, dx=\int_{n \delta }^\infty e^{-y^2}\, dy=\frac{1}{2} \sqrt{\pi } \text{erfc}\left({n\delta }\right)$$ For large values of $z$ (see here)
$$\text{erfc}(z)=\frac{e^{-z^2}}{\sqrt \pi}\left(\frac{1}{z}-\frac{1}{2 z^3}+O\left(\frac{1}{z^5}\right) \right)$$ making
$$I\sim \frac 12 \frac{e^{-n^2\delta^2}}{n \delta}$$
A: The key is that the exponential is decaying really fast. Since $x\geq \delta$ in the interval of integration you get
$$
0\leq \int_\delta^{+\infty} ne^{-n^2x^2}\,dx
\leq \frac{1}{\delta}\int_\delta^{+\infty}nxe^{-(nx)^2}\,dx
=\frac{e^{-(n\delta)^2}}{2n\delta}.
$$
Now the squeeze theorem gives the result.
