Let $f_n(x)=n(\sin x)^n \cos x$.
Show that the sequence of functions $f_n$ converges to $0$ uniformly on any interval of the form $[0,a]$ where $a<\pi /2$.
Show that, for any continuous function $g\in C([0,\pi /2])$
$$\lim_{n\to \infty} \int^{\pi/2}_0f_n(x)g(x){\rm d}x=g(\pi/2)$$
For 1. it is clear point-wise limit is $0$. Hopefully, we could show it converge uniformly too, but i have not tried.
For 2. My idea is to write $n\to \infty \frac{ \int^{\pi/2}_0 (\sin x)^n \cos x g(x)}{ \int^1_0x^{n-1}{\rm d}x}{\rm d}x$
Since numerator and denominator both goes to $0$ as $n \to \infty$, i can use L'Hopital rule, which give me $0$ but it should $g(\pi/2)$, Where am I wrong? please correct my idea.
For $a)$ you want to show that the whole expression is less than than $n c^n$, where $c=\sin(a)$.
In $b)$, take $u=\sin(x)$ to turn the problem into
$$\lim_{n \to \infty} \int_0^1 n u^n g \left ( \arcsin(u) \right ) du$$
Now the idea is to split the region of integration into $[0,1-\delta]$ and $[1-\delta,1]$ for an appropriate $\delta$, use uniform convergence to estimate the first part and continuity of $u \mapsto g \left ( \arcsin(u) \right )$ to estimate the second part.
One way to get intuition for this is to consider
$$\int_0^1 n x^n f(x) dx$$
where $f$ is continuously differentiable, and integrate by parts. This is how I came up with the solution to a previous qual problem (where we were not given that $f$ was continuously differentiable), and then reverse engineered the derivation above.