Taking the limit of an integral expression Hi guys I was doing the following question:-

let a,b ∈ $\mathbb{R}$ with a &lt; b show the following

$$\lim_{n\to \infty}\int^a_b\frac{((b-a){e^x}^2)}{(b-a)n^2{e^x}^2+b^2} dx = 0 $$

so i started taking the lim as $\lim{n\to \infty}$ which gives the following:-

$$\lim_{n\to \infty}\frac{((b-a){e^x}^2)}{(b-a)n^2{e^x}^2+b^2}  $$

Since i am trying to evaluate the limit I decided to remove the top expression since its a constant and ended up getting

$$(b-a){e^x}^2\lim_{n\to \infty}\frac{1}{(b-a)n^2{e^x}^2+b^2}  $$

Now looking at the denominator expression can i directly substitute n as infinity

therefore this gives the following:-

$$(b-a){e^x}^2\lim_{n\to \infty}\frac{1}{(b-a)(\infty)^2{e^x}^2+b^2}  $$

Can i take this expression to be $\frac{1}{\infty} = 0 $

hence i get the following $$(b-a){e^x}^2(0) = 0 $$

Therefore when i go to the next part of the question to evaluate the integral well = 0 since the above shows that when the limit is evaluated f(x) = 0. I am not sure if i evaluated this question correctly when looking at the limit aspect and so hoping so one can confirm this for me.
You need, for example, Lebesgue Dominated Convergence Theorem to guarantee the swiping of the integral with the limit to move on.

Or you can do it in the following way, because the integrand has no singularity:
\begin{align*}
\dfrac{(b-a)e^{x^{2}}}{(b-a)n^{2}e^{x^{2}}+b^{2}}\leq\dfrac{(b-a)e^{b^{2}}}{(b-a)n^{2}e^{a^{2}}+b^{2}}\leq\dfrac{(b-a)e^{b^{2}}}{(b-a)n^{2}e^{a^{2}}}=\dfrac{e^{b^{2}}}{e^{a^{2}}}\dfrac{1}{n^{2}}.
\end{align*}
Taking integral both sides, the right sided is just $(b-a)\dfrac{e^{b^{2}}}{e^{a^{2}}}\dfrac{1}{n^{2}}$ and this goes to zero, now you use Squeeze Theorem to conclude that the integral also goes to zero.
$0$ is  the correct limit you can get it from the fact that the integrand lies between $0$ and $\frac 1 {n^{2}}$ because $b^{2} \geq 0$. Hence the integral itself lies between  $0$ and  $\frac {b-a} {n^{2}}$. You don't have to evaluate the integral or apply a theorem like DCT.