Evaluate $$\lim_{n\rightarrow \infty} \int_0^1 \frac{n^{(3/2)}x}{1+n^2x^2} \,\mathrm{d}x \,.$$
Solution: Let $$f_n(x) = \frac{n^{(3/2)}x}{1+n^2x^2} \,.$$ It is clear that $f_n \rightarrow 0$ regardless of $x$ as $n$ grows large, and thus that $f_n$ converges to zero pointwise.
Now we have $$\frac{n^{(3/2)}x}{1+n^2x^2} = \sqrt{n}\frac{nx}{1+n^2x^2} \,. $$ Let $t = nx$, so that the above expression becomes $$\frac{1}{\sqrt{x}}\sqrt{t}\frac{t}{1+t^2} \,,$$ and define $$h(t) = \frac{t^{3/2}}{1+t^2} \,.$$ Since $h(0) = 0$ and $h(t) \rightarrow 0$ as $t \rightarrow \infty $, and $h(t)$ has no other possible sources of discontinuity, it follows that $h(t)$ is bounded. It thus follows that $$\frac{n^{(3/2)}x}{1+n^2x^2} \leq \frac{1}{\sqrt{x}}\times M$$ for some real positive constant $M$. Now since $1 / \sqrt{x} \in \mathscr{L}^1$ on $[0,1]$, we have the sequence is bounded above by an $\mathscr{L}^1$ function, and the dominated convergence theorem applies. It follows that the original integral tends to zero.
HINT: your integral is given by $$\frac{\log \left(n^2+1\right)}{2 \sqrt{n}}$$ Can you compute the limit? the searched limit is $0$