Need to find maximum in order to prove that the function converges uniformly. So, here is the function: $f_n = \sqrt{n}\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$
I found that the limit for it is $0$. Now I want to show whether it converges uniformly. So first I tried to find a maximum.
$$f'_n(x) = \frac{1}{2}\sqrt{n}\left(\frac{1}{\sqrt{x+\frac{1}{n}}}-\frac{1}{\sqrt{x}}\right) = \\ \frac{1}{2}\sqrt{n}\left(\frac{\sqrt{x}-\sqrt{x+\frac{1}{n}}}{\sqrt{x}\sqrt{x+\frac{1}{n}}}\right)$$
Then I solve for the x at denominator.
$$\sqrt{x}-\sqrt{x+\frac{1}{n}}=0 \Rightarrow x -2\sqrt{x}\sqrt{x+\frac{1}{n}}+x+\frac{1}{n}= 0 \\ 2\sqrt{x}\sqrt{{x+\frac{1}{n}}}=\frac{1}{n}+2x \Rightarrow 2x^2+2x\frac{1}{n}=\frac{1}{n^2}+4x\frac{1}{n}+4x^2 \\ 2x^2+2x\frac{1}{n}+\frac{1}{n^2}=0 \ \ x_1 = \frac{-2\frac{1}{n}+ \sqrt{\frac{4}{n^2}-\frac{8}{n^2}}}{6} = -\frac{1}{3n}+\frac{1}{3n}=0$$
$$x_2 = -\frac{2}{3n}$$
I get very clumsy $x_1$ so I gather I have made some mistake all along. Can someone help?

Your sequence of functions converges pointwise to $0$ on $\mathbb{R}_+^*$. Thus, it converges uniformly to zero if $||f_n||_{\infty}= \sup_{x}|f_n(x)| \to 0$. But $f_n(x) \to 1$ as $x \to 0$, so $||f_n||_{\infty}\geqslant 1$ for all $n$ and $(f_n)$ does not converge uniformly
