Question about change of variables in limits of integrals using dominated convergence Hello my question is really inspired by the following example, but also many others which are similar. Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $

Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $
where $f:[-1,1] \to \mathbb{R}$ is a continuous function.

When we have a limit of an integral to compute, and we change variables to get it to a more tractable form, and the change of variables is a function of n and x, say here its $t= \sqrt{n} x$, thereby eliminating the $nx^2$ from that term and sort of covering it by t, and then we take the limit of the whole function as $ n\to \infty$, aren't we assuming t is independent of n when it is dependent on n?
Thanks in advance.
 A: 
Upon enforcing the substitution $\displaystyle t=\sqrt n x$, for $\displaystyle  x\in [-1,1]$, $\displaystyle  t\in [-\sqrt n,\sqrt n]$ becomes a "dummy" index that spans $[-\sqrt n,\sqrt n]$.  
It is, therefore, not dependent on $n$, but rather it spans a domain that depends on $\displaystyle  n$.

Proceeding, we have
$$\begin{align}
\sqrt{n}\int_{-1}^1 e^{-\frac12nx^2}f(x)\,dx&=\int_{-\sqrt n}^{\sqrt n} e^{-\frac12t^2}f(t/\sqrt n)\,dt\\\\
&=\int_{-\infty}^\infty e^{-\frac12t^2}f(t/\sqrt n))\xi_{[-\sqrt n,\sqrt n]}(t)\,dt
\end{align}$$
Inasmuch as $f\in C[-1,1]$, it is bounded there.  Then, we have
$$\left|e^{-\frac12t^2}f(t/\sqrt n))\xi_{[-\sqrt n,\sqrt n]}(t)\right|\le ||f||_{\infty}e^{-\frac12 t^2}\in L^1$$
where $||f||_{\infty}=\sup_{x\in [-1,1]}|f(x)|$.
Hence, we have 
$$\begin{align}
\lim_{n\to \infty}\sqrt{n}\int_{-1}^1 e^{-\frac12nx^2}f(x)\,dx&=\int_{-\infty}^\infty \lim_{n\to \infty}\left(e^{-\frac12t^2}f(t/\sqrt n))\xi_{[\sqrt n,\sqrt n]}(t)\right)\,dt\\\\
&=f(0)\int_{-\infty}^\infty e^{-\frac12t^2}\,dt\\\\
&=\sqrt{2\pi }f(0)
\end{align}$$
And we are done!
