# Compute $\lim_{n\to \infty}\int_0^{1/n}\frac{n}{1+n^2x^2+n^6x^8}dx$

The integrand decreases when $n\ge 1$ but the statement of the monotone convergence theorem requires that the sequence is increasing. Fatou's lemma doesn't apply since it will only give a lower bound of $0$. That leaves the dominated convergence theorem (DCT) , but to use it I'd have to find an integrable upper bound for $\frac{n}{1+n^2x^2+n^6x^8}$ but I can't find a better upper bound than $f(x)=n$ and I'm not sure if such a function is allowed as an upper bound when using DCT since it depends on $n$.

It seems clear that the integrand and the region of integration $[0,1/n]$ tend to $0$ but I don't see how to show it formally.

## 2 Answers

Let $u=nx$ then the given integral becomes

$$\int_0^1\frac{du}{1+u^2+\frac1{n^2}u^8}$$ and the function which dominates is $\phi(u)=\frac1{1+u^2}$. Can you take it from here using the DCT?

Hint. You may perfom the change of variable $u=nx$, giving $ndx=du$ to get $$\int_0^{1/n}\frac{n}{1+n^2x^2+n^6x^8}dx=\int_0^{1}\frac{1}{1+u^2+\dfrac{u^8}{n^2}}du\leq\int_0^{1}\frac{1}{1+u^2}du$$ and then easily use the dominated convergence theorem.