If a function does not converge point-wise, then it doesn't converge uniformly as well, right?

Question

Here is the function: $f_n(x) = \sqrt{\sin^2 x + n^{-4}}, \ \ x \in R$. If $x \neq \frac{\pi}{2}+2\pi k \text{ or } x \neq 2\pi k, \ \ k \in Z$ then the function doesn't converge at all even point wise, am I right? So, how do I show formally that it doesn't converge?

Answer 1

Define $f(x)=|\sin x|$. Hence $$|f_n(x)-f(x)|{=\left|\sqrt {\sin^2 x+n^{-4}}-|\sin x|\right| \\=\left|{(\sqrt {\sin^2 x+n^{-4}})^2-|\sin x|^2\over \sqrt {\sin^2 x+n^{-4}}+|\sin x|}\right| \\=\left|{{\sin^2 x+n^{-4}}-\sin^2 x\over \sqrt {\sin^2 x+n^{-4}}+|\sin x|}\right| \\=\left|{n^{-4}\over \sqrt {\sin^2 x+n^{-4}}+|\sin x|}\right| \\={n^{-4}\over \sqrt {\sin^2 x+n^{-4}}+|\sin x|} \\\le {n^{-4}\over \sqrt {\sin^2 x+n^{-4}}} \\\le {n^{-4}\over \sqrt {n^{-4}}} \\= {n^{-4}\over {n^{-2}}} \\={1\over n^2} \\<\epsilon }$$ hence by choosing $n>\sqrt{1\over \epsilon}$, we can have $f_n(x)$ arbitrarily close to $f(x)$ at any point, hence $f_n(x)$ converges to $f(x)$ uniformly.