How do you show that $\lim_{x \rightarrow a} f(x)=0$? This is a simple proof of Real Analysis.
Show that $\lim_{x \rightarrow a} f(x)=0$, given $|f(x)|\leq g(x)$, $\forall x\in\mathbb R$ and $\lim_{x \rightarrow a} g(x)=0$.
How do you show that $\lim_{x \rightarrow a} f(x)=0$?
I tried by myself  $\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|x-a|<\delta\implies |g(x)|<\epsilon.$ Since $g(x) \leq f(x)\leq g(x)$ 
By squeeze theorem. I think  $\lim_{x \rightarrow a} g(x)=0$ 

You know $\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|a-x|<\delta\implies |g(x)|<\epsilon.$ Since $|f(x)|\leq g(x)\leq |g(x)|$ it follows
$$\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|a-x|<\delta\implies |f(x)|\leq |g(x)|<\epsilon.$$
