Dominated convergence theorem calculating integral How do you evaluate $$\lim_{n \to \infty} \int_{0}^{1}\frac{n\cos(x)}{1+n^2x^{3/2}}\, dx\;\;?$$
I know to use the Dominated convergence theorem, however I cannot find a function that will dominate $$\frac{n \cos(x)}{1+n^2x^{3/2}}.$$
Any help will be much appreciated.
 A: $|\cos(x)| \leq 1$, and $|1 + n^2 x^{3/2}| \geq |2nx^{3/4}|$, so
$$\begin{aligned}
\left|\frac{n \cos(x)}{1+n^2x^{3/2}}\right|
&\leq \left|\frac{n}{2nx^{3/4}}\right| \\
&= \left|\frac{1}{2x^{3/4}}\right|\\
\end{aligned}$$
which is integrable over $[0,1]$. To prove the key inequality $|1 + n^2 x^{3/2}| \geq |2nx^{3/4}|$, let $y = x^{3/4}$, so $y^2 = x^{3/4}$. Then
$$0 \leq (1 - ny)^2 = 1 + n^2 y^2 - 2ny$$
so
$$2ny \leq 1 + n^2 y^2$$
and therefore
$$2nx^{3/4} \leq 1 + n^2 x^{3/2}$$
Both sides are nonnegative since $x \geq 0$, so the absolute values change nothing:
$$|2nx^{3/4}| \leq |1 + n^2 x^{3/2}|$$
A: Consider a change of variables $y=nx$ (and so we have $\mathop{dy}= n \mathop{dx})$. This allows us to alter $$\frac{n\cos(x)}{1+n^2x^{3/2}}\mathop{dx}$$ into:
$$\dfrac{n\cos \left(\frac{y}{n}\right)}{1+n^2\left(\frac{y}{n}\right)^{3/2}}\mathop{dx}=\dfrac{\cos \left(\frac{y}{n}\right)}{1+n^{1/2}y^{3/2}}\mathop{dy}$$
Consider the dominating function $f(y)=\dfrac{1}{1+y^{3/2}}$
The result then follows from DCT.
A: Here is an answer that completes the above @Sujaan Kunalan idea:

