If $f_n\to f$ uniformly and $\lim_{x \to \infty}f_n(x)=0$ for all $n$. Then $\lim_{x \to \infty}f(x)=0$

Question

If $f_n\to f$ uniformly and $\lim_{x \to \infty}f_n(x)=0$ for all $n$. Then $\lim_{x \to \infty}f(x)=0$. Is this statement true?

My try:

Pick $\epsilon>0$ then $\exists N$ s.t $|f_n-f|<\epsilon$ $\forall x$. Now it is tempting to simply take the limit of both sides where $x$ goes to infinity but i am not sure if that is justified. Instead, we know there exists $M$ s.t if $M<x$ then $|f_N(x)|<\epsilon$ Thus by reverse triangle inequality and large enough $x$ we have $|\epsilon-f(x)|<\epsilon$ Hence $|f(x)|<2\epsilon$ Thus $\lim_{x \to \infty}|f(x)|=0$ and so $\lim_{x \to \infty}f(x)=0$ is this correct?

Answer 1

The line that $|\epsilon-f(x)|<\epsilon$ seems not correct.

I will do it in the following way, not quite favorable though.

We have $f(x)\leq\epsilon+f_{n}(x)$ for all $x$ and $n\geq N$, in particular, $f(x)\leq\epsilon+f_{N}(x)$, taking limit supremum both sides, we have $\limsup_{x\rightarrow\infty}f(x)\leq\epsilon$.

We also have $f_{N}(x)\leq\epsilon+f(x)$, taking limit infimum both sides, we have $0\leq\epsilon+\liminf_{x\rightarrow\infty}f(x)$, so $-\epsilon\leq\liminf_{x\rightarrow\infty}f(x)\leq\limsup_{x\rightarrow\infty}f(x)\leq\epsilon$. Taking $\epsilon\rightarrow 0^{+}$, we have $0\leq\liminf_{x\rightarrow\infty}f(x)\leq\limsup_{x\rightarrow\infty}f(x)\leq 0$, and hence $\lim_{x\rightarrow\infty}f(x)=0$.