How to show differentiability implies continuity for functions between Euclidean spaces

A function $f: \mathbb{R^n} \to \mathbb{R^m}$ is differentiable at $a$ if there exists a linear map $ \lambda: \mathbb{R^n} \to \mathbb{R^m}$ such that

$$\lim_{h \to 0} \frac{\|f(a+h) - f(a) - \lambda(h)\|}{\|h\|} = 0$$

So clearly, if $f$ is differentiable at $a$ then $\lim_{h \to 0} f(a+h) - f(a) - \lambda(h) = 0$, but where do you proceed from here?

More explicitly:

The limit shows that for any $\epsilon>0$, there exists a $\delta>0$ so that if $\|h\| < \delta$, then $\|f(a+h) - f(a) - \lambda(h)\| \leq \epsilon \|h\|$. $\lambda$ is continuous, hence bounded, so we have $\|\lambda(h)\| \leq K \|h\|$, for some $K$.

Then we have $\|f(a+h) - f(a)\| \leq \|f(a+h) - f(a) - \lambda(h)\| + \|\lambda(h)\|\leq (\epsilon + K ) \|h\|$.

Now let $\eta >0$, then if $\|h\| < \min(\delta, \frac{\eta}{\epsilon+K})$, we have $\|f(a+h) - f(a)\| < \eta$, which shows that $f$ is continuous at $a$.

Note that $\lambda$ is linear, so $\lim\limits_{h\to 0} \lambda(h)=0$. Thus $\lim\limits_{h\to 0} f(a+h)-f(a)=0$, which is equivalent to $\lim\limits_{x\to a}f(x)=f(a)$, one of the definitions of continuity (of course all the definitions are equivalent).