if $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ is differentiable at $a\in\mathbb{R}^n$ then it is continuous at $a$. Prove that if $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ is differentiable at $a\in\mathbb{R}^n$ then it is continuous at $a$.

My attempt:
My intuition tells me that to show continuous, I need to show that $lim_{x \rightarrow a} f(x)= f(a)$.
By using the book "calculus on manifolds,"....
Let $f: \mathbb{R}^n \rightarrow \mathbb {R}^m$ be differentiable at $a \in \mathbb{R}^n$, then we can say that there exists a linear transformation $\lambda: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $$ lim_{h \rightarrow0}= \frac {||f(a+h)-f(a)-\lambda(h)||}{||h||}=0$$
I am new to this and I am not sure where to go from here.
 A: Check out this question I posted :
Question
You should be able to adapt it with your more general definitions!
A: Assume $f:\mathbb{R}^n\to \mathbb{R}^m$ is differentiable, then there exists a linear transformation $\lambda: \mathbb{R}^n\to \mathbb{R}^m $ such that $$\tag{1}\lim_{h\to 0}\frac{|f(a+h)-f(a) - \lambda (h)|}{|h|}=0.$$
Let $A=(a_{ij})$ be the Jacobian of $\lambda$, $\alpha = \max{|a_{ij}|}$, and $M=\sqrt{mn}\alpha$, then$^*$
$$\tag{2}|\lambda(h)|\leq M|h|$$ for all $h \in \mathbb{R}^n$.
We wish to show $\lim_{x \to a} f(x)=f(a)$. We can use $(2)$ to derive the inequality $$\frac{|f(x)-f(a) - \lambda(x-a)|}{|x-a|}\geq \frac{|f(x)-f(a)| - |\lambda(x-a)|}{|x-a|} \geq \frac{|f(x)-f(a)|}{|x-a|} - M.$$ Let $ \epsilon > 0 $ be given. Let $h=x-a$ in $(1)$, then for any $ \epsilon_1>0$ we can find $\delta >0$ such that $$\epsilon_1 > \frac{|f(x)-f(a) - \lambda(x-a)|}{|x-a|}$$ for all $x\in \mathbb{R}^n$ satisfying $|x-a|<\delta$. Since $\epsilon_1$ was arbitrary, we can choose the pair $(\epsilon_1, \delta)$ small enough that $$\epsilon > \delta( \epsilon_1 + M) > |x-a| (\epsilon_1 + M) > |f(x)-f(a)|.$$ $ \square$
$*$ Proof of (2):
By Cauchy-Schwarz, for any set of real numbers $x_1, ... , x_k$: $$\tag{3}(x_1+...+x_k)^2\leq k(x_1^2+...+x_k^2).$$
Consider
$$y^i= a_{i1} (h^1)+...+a_{in}(h^n),$$  then $$(y^i)^2= |(y^i)|^2\leq \left( |a_{i1} (h^1)|+...+|a_{in}(h^n)|\right)^2 \leq \alpha^2\left( \sum_{j=1}^{n} (h^j) \right)^2\leq n\alpha^2|h|^2, $$ where the last inequality follows from $(3)$. Thus
$$|\lambda(h)| ^2= \sum_{i=1}^{m} (y^i)^2 \leq mn\alpha^2 |h|^2= M^2 |h|^2.$$
$\square$
