# Taking the limit of an integral expression

Hi guys I was doing the following question:-

let a,b ∈ $$\mathbb{R}$$ with a < 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 LaTeX says the oddly formatted exponentials are $(\mathrm{e}^x)^2$, not $\mathrm{e}^{x^2}$. Are you sure that is correct? – Eric Towers Oct 29 '19 at 5:33
• yes thats correct what i typed is what the question has in it – Amir Oct 29 '19 at 5:36
• I ask because the usual way to write $(\mathrm{e}^x)^2$ is "$\mathrm{e}^{2x}$". – Eric Towers Oct 29 '19 at 5:36
• yeah i understand but the question has it as i typed it so yeah thats the correct notation that i provided. – Amir Oct 29 '19 at 5:40

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.

• I was not introduced to Lebesgue Dominated convergence so thats why i did it like that. – Amir Oct 29 '19 at 5:33
• Anyway, the second one uses elementary analysis, no hard theorem anyway. – user284331 Oct 29 '19 at 5:34
• so based on my answer is that completely inaccurate or incorrect rather based on my approach of the limit to then go onto to the integral? – Amir Oct 29 '19 at 5:38
• Actually yours is correct, just without more reasoning to support. You implicitly swiped the integral and the limit, in this case, it goes through. – user284331 Oct 29 '19 at 5:40
• OP seems to have clarified in comments that the integrand you use is not OP's integrand. – Eric Towers Oct 29 '19 at 5:42

$$0$$ 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.

• i was considering this $$(b-a){e^x}^2\lim_{n\to \infty}\frac{1}{(b-a)n^2{e^x}^2+b^2} \leq \frac{1}{n^2}$$ would be accurate to say or is that what your saying ? Oh btw my evaluation of the expression was done properly? – Amir Oct 29 '19 at 5:28