Investigate uniform convergence of:
$$f_n(x) = n\left( 1- \sqrt{1 + \frac{x}{n}}\right)$$
in the interval $E = (0,1)$
Attempt:
First we find $f(x)$:
$$f(x) = \lim\limits_{n\to \infty}f_n(x) \sim n\left( 1- \left(1 + \frac{x}{2n}\right)\right) = -\frac{x}{2}$$
In order for $f_n(x)$ to uniformly converge in $E$, is necessary and sufficient that $$\lim\limits_{n\to \infty} \mathrm{sup}_{x\in E}|f_n(x) - f(x)| = 0$$
What should I do in this part?
First I was trying to find some extrema of $\ f_n(x)$ but I got a $x_{extrema}$ that is not in $E$. Then I just considered that $\mathrm{sup}\ f_n(x) = 0$ but in that case $|f_n(x) - f(x)| \neq 0$.
Finally I was considering taking the Taylor series of $|f_n(x) - f(x)|$, that gives $\left|-\frac{x}{2}+\frac{x}{2}\right|= 0$, but I feel I'm just making up the whole thing in order to get the result I wanted (since I'm essentially taking a Taylor series of $f_n(x)$ twice, which obviously gives $0$ in any case).
Consider $g_n(x) = |f_n(x) - f(x)|$ on the set $E = (0,1)$. This function is continuous on $E$ for all $n \in \mathbb{N}$.
When the stationary points are outside the set of interest, then the $\sup$ and the $\inf$ can be found on the border of the set (look at this).
For $x=0$, we have that $g_n(0) = 0$, while for $x=1$, we have that:
$$g_n(1) = \left|n\left(1-\sqrt{1+\frac{1}{n}}\right) + \frac{1}{2}\right|.$$
Note that $g_n(1) \geq 0$ for all $n$, and hence $$\mathrm{sup}_{x\in E} = g_n(1) = \left|n\left(1-\sqrt{1+\frac{1}{n}}\right) + \frac{1}{2}\right|.$$
Now, it is easy to show that
$$\lim\limits_{n\to \infty} \mathrm{sup}_{x\in E}|f_n(x) - f(x)| = 0,$$
since
$$\lim\limits_{n\to \infty} n\left(1-\sqrt{1+\frac{1}{n}}\right) = -\frac{1}{2}.$$