I have the following exercise:
Consider the following sequences of functions $$f_n : [0,1] \to \mathbb R \,,\ f_n(x) := \frac{\sqrt x (1-x)}{1+nx}$$ $$g_n : [0,1] \to \mathbb R \,,\ g_n(x) := \sqrt n f_n(x)$$ Study the pointwise and uniform convergence in both cases. Eventually, calculate the limits: $$\lim_{n \to \infty}\int_0^1 f_n(x) \mathrm dx \quad \text{and} \quad \lim_{n \to \infty}\int_0^1 g_n(x) \mathrm dx$$
I have come up with a partial solution by myself, but I'd like to see if there is some "trick" to get all straight.
Here is may attempt.
First of all, $\lim_{n \to \infty} f_n(x) = 0$ for every $x \in [0,1]$, and the same does $g_n$. For uniform convergence, I have seen if the functions have some nice upper bounds: the point of maximum for $f_n$ is $$x_n := \frac{-n-3+\sqrt{(n+1)(n+9)}}{2n}$$ $f_n \left(x_n\right)$ is not nice at all, but the is a nice asymptotic estimation: $$f_n\left(x_n\right) \sim \frac 1 {\sqrt n} \quad \text{for } n \to \infty$$ Owing to this, I can conclude that $f_n \to 0$ uniformly too. In contrast, $g_n$ does not because $\lim_{n \to \infty} g_n\left(x_n \right) = 1$.
About the sequence of integrals, I can just conclude the the first of two limits is $0$, by uniform convergence of $f_n$ to $0$. What about $g_n$?
$f_n\to0$ uniformly on $[0,1]$: For $x\in[0,1/n]$ we have $|f_n(x)|\leq \sqrt x\leq \frac1{\sqrt n}$; for $x\in[1/n,1]$ we have $|f_n(x)|\leq\frac{\sqrt x}{nx}\leq\frac1n\frac1{\sqrt x}\leq \frac1{\sqrt n}$. Hence $|f_n(x)|\leq \frac1{\sqrt n}$ for all $x\in[0,1]$ and thus $f_n\to0$ uniformly on $[0,1]$.
$g_n(x)=\frac{\sqrt{nx}(1-x)}{1+nx}$ is NOT uniformly convergent: Note that $g_n(x)\to0$ for all $x\in[0,1]$ but $$g_n\left(\frac1n\right)=\frac12\left(1-\frac1n\right)\to\frac12.$$
$\lim_{n\to\infty}\int_0^1g_n(x)\,dx=0$: We have $$0\leq\int_0^{1/n}g_n(x)\,dx\leq \int_0^{1/n}\sqrt{nx}\,dx=\frac2{3n},$$ and $$0\leq\int_{1/n}^1g_n(x)\,dx\leq\int_{1/n}^1\frac{\sqrt{nx}}{nx}\,dx=\frac1{\sqrt n}\int_{1/n}^1\frac1{\sqrt x}\,dx<\frac2{\sqrt n}.$$ Hence $$0\leq \int_0^1g_n(x)\,dx\leq\frac2{3n}+\frac2{\sqrt n}\to0.$$