How can we evaluate this with Lebesgue dominated convergence theorem?

Question

$$\lim\limits_{n\to\infty}\int\limits_0^1\dfrac{x^2}{x^2+(1-nx)^2}dx$$ shall we add and remove$$\lim\limits_{n\to\infty}\int\limits_0^1\dfrac{x^2+(1-nx)^2-(1-nx)^2}{x^2+(1-nx)^2}dx$$ and going this way $$\lim\limits_{n\to\infty}\int\limits_0^11-\dfrac{(1-nx)^2}{x^2+(1-nx)^2}dx$$

Answer 1

Theorem (Dominated Convergence)

Let $(X,\Sigma,\mu)$ be a measure space and $f_n:X\to\mathbb{R}$ be a succession of functions all measurable with respect to $\Sigma$ and the real Borel sets. If $f_n\to f$ pointwise (or pointwise almost everywhere) and $|f_n|\leq g$ and $|f|\leq g$ -- or both of these hold almost everywhere --, where $g$ is an integrable function $g:X\to\mathbb{R}$, then: $$\lim_{n\to\infty}\int_Xf_n\mathrm{d}\mu=\int_Xf\mathrm{d}\mu.$$

Proof

See Wikipedia.

Application to our case

In our case, $(X,\Sigma,\mu)=([0,1],\mathcal{B}([0,1]),\lambda)$, where $\lambda$ is the Lebesgue measure. We know that if both the Riemann Integral and the Integral with respect to $\lambda$ exist, then they are equal. We have to evaluate the LHS of the theorem's equality. We have: $$f_n(x)=\frac{x^2}{x^2+(1-nx)^2}.$$ Let's try to find that $g$: $$|f_n(x)|\leq\frac{x^2}{x^2}=1,$$ and $1$ is integrable, so our $g(x)=1$ for all $x\in X$. By the above theorem, the limit passes under the integral, so we now need the pointwise limit of $f_n$. $$\begin{align*} f_n(x)={}&\frac{x^2}{x^2+(1-nx)^2}=\frac{x^2}{x^2+n^2x^2-2nx+1}=\frac{x^2}{(n^2+1)x^2-2nx+1}\sim{} \\ {}\sim{}&\frac{x^2}{(1+n^2)x^2}=\frac{1}{n^2+1}\to0. \end{align*}$$ So the pointwise limit -- the $f$ in the above theorem -- is 0, meaning the limit we have to evaluate is 0.

Exercise

Use the Dominated Convergence Theorem in a similar way to establish that the subtracted term in your way of rewriting the integral in the question has an integral tending to 1.

Answer 2

For all $n$, $$x^2+(1-nx)\geq x^2\implies \frac{x^2}{x^2+(1-nx)^2}\leq 1\in L^1(0,1)$$

therefore you can permute limit and integral and you finally find $$\lim_{n\to\infty }\int_0^1\frac{x^2}{x^2+(1-nx)^2}dx=\int_0^1\underbrace{\lim_{n\to\infty }\frac{x^2}{x^2+(1-nx)^2}}_{=0}dx=0.$$