I realize this question has answers all over the place, but I'm still lost. Specifically, I want to prove that given a function $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ that is differentiable at the point $c$, it follows that $f$ is continuous at $c$. The definition of differentiable that I'm using is $f$ is differentiable at $c$ if there exists a linear map $L:\mathbb{R}^n\rightarrow\mathbb{R}^m$ such that $$\lim_{h\rightarrow0}\frac{\|f(c+h)-f(c)-L(h)\|_{\mathbb{R}^m}}{\|h\|_{\mathbb{R}^n}}=0.$$ I understand that the setup is something along these lines:
By application of the triangle inequality, we have $\|f(c+h)-f(c)\|=\|h\|\cdot\frac{\|f(c+h)-f(c)\|}{\|h\|}\leq\|h\|\cdot\frac{\|f(c+h)-f(c)-L(h)\|}{\|h\|}+\|L(h)\|$
But then all the proofs I've found seem to stop here or shortly after. Sure, it's clear that the RHS inequality goes to $0$ as $h\rightarrow0$. But why is it that $\lim_{h\rightarrow0}\|f(c+h)-f(c)\|=0$ implies that the function is continuous? I need to show that $\lim_{h\rightarrow c}f(h)=f(c)$, correct?
By application of the triangle inequality, we have $$||F(c+h)-F(c)||=||h||\cdot\frac{||F(c+h)-F(c)||}{||h||}\leq||h||\cdot\frac{||F(c+h)-F(c)-L(h)||}{||h||}+||L(h)||.$$ Then, because $L$ is a linear map, as $h\rightarrow 0$, the RHS of the inequality goes to $0$. That is, $$\lim_{h\rightarrow0}||F(c+h)-F(c)||=0.$$ This is true if and only if $$\lim_{h\rightarrow0}F(c+h)=F(c).$$ Then, denote $t=c+h$. It follows that $$\lim_{t\rightarrow c}F(t)=F(c),$$ and hence $F$ is continuous at $c$.