<s>[INST] How to find the following limit $\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx$ Does the following limit exists
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx
\end{align}
Observe that if we do 
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \le  \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{e^{-\frac{x^2}{2}}} dx  = \lim_{a \to \infty} e^{a^2}=\infty
\end{align}
or  another upper bound is 
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \le  \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}} dx  = \lim_{a \to \infty} a^2 \sqrt{2 \pi}=\infty
\end{align}
For lower bounds  we get
\begin{align}
\lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}e^{-\frac{(x-a)^2}{2}}+e^{-\frac{x^2}{2}}} dx \ge  \lim_{a \to \infty} \int \frac{e^{-(x-a)^2}}{\frac{1}{a^2}+1} dx = \lim_{a \to \infty} \int e^{-(x-a)^2}dx= \lim_{a \to \infty}\frac{\sqrt{2 \pi}}{\frac{1}{a^2}+1} =\sqrt{2 \pi}
\end{align}
Yet on the other had we get
\begin{align}
 \int  \lim_{a \to \infty} \frac{e^{-(x-a)^2}}{\frac{1}{a^2}+e^{-\frac{x^2}{2}}} dx =0
\end{align}
This together with the lower bound suggests that switching limit and integration is not allowed. So, I don't think that DCT or MCT apply? 
Can someone explain how to approach this problem?  
 [/INST] Call your integral $I_a.$ Let $x=y+a$ in the integral, then reduce the interval of integration to $[0,1].$ Then we have
$$\tag 1 I_a > \int_0^1 \frac{e^{-y^2}}{(e^{-y^2/2})/a^2 + e^{-(y+a)^2/2}}\, dy > \int_0^1 \frac{e^{-y^2}}{(e^{-y^2/2})/a^2 + e^{-a^2/2}}\, dy$$ $$ = a^2\int_0^1 \frac{e^{-y^2}}{e^{-y^2/2} + a^2e^{-a^2/2}}\, dy  $$
Now $a^2e^{-a^2/2} \to 0$ as $a\to \infty.$ So a very simple application of the DCT shows the last integral $\to \int_0^1 e^{-y^2/2}\, dy.$ Therefore $I_a$ is at least a positive constant times $a^2$ for large $a,$ hence $I_a \to \infty.$  
</s>