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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.

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    you want to prove $\lim_{x\to a} f(x) = f(a)$ well this is equal to proving that $\lim_{x\to a} f(x) - f(a)=0$ . There should be a second definition of differentiability, can you see how it is similar to this expression? – 2017-01-26

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Check out this question I posted : Question

You should be able to adapt it with your more general definitions!