Let $f\colon A\subseteq \mathbb{R^n}$ $\rightarrow$ $\mathbb{R}^m$ and $h:A \subseteq \mathbb{R^n} \rightarrow \mathbb{R^{\geq0}}$ Assume $ > 0 \leq |f(x)| \leq h(x)$ for $x$ in a punctured ball of $a$. Show that if $h(x) \rightarrow 0$ as $x \rightarrow a$ then $f(x)\rightarrow 0$ as $x \rightarrow a$.
My Proof:
Assume $h(x) \rightarrow 0$ as $x \rightarrow a$. Let $\epsilon >0$. So $\exists$ $\delta>0$ such that whenever we have $0 < |x-a| < \delta$ we have then $|h(x)| < \epsilon$. Because $|f(x)|,h(x)\geq 0$ it follows that if I set $\delta'=\delta$ then whenever $0<|x-a|<\delta'$ we have $||f(x)||=|f(x)| <\epsilon$.
Is this proof correct? I would very much like feedback.
Let $\epsilon > 0$
Since $h(x) \to 0$ when $x \to a$ this means that for our particular $\epsilon$ there $\exists \delta$ s.t. whenever $|x - a| < \delta \Rightarrow |h(x)| < \epsilon$
Let $\delta' = \delta$ then $|x-a| < \delta \Rightarrow$ eventually $x \in B_{\delta}(a) -\{a\} \Rightarrow 0 \leq |f(x)| \leq h(x) \leq \epsilon \Rightarrow |f(x)| \leq \epsilon$.
Thus $\forall \epsilon > 0$ there $\exists \delta'$ s.t $|x-a| < \delta' \Rightarrow |f(x)| < \epsilon$ and thus $f(x) \to 0$ when $x \to a$