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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?

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3 Answers 3

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 $a$.

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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).

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Why is this step justified? –  WacDonald's Jul 19 '12 at 2:07
@WacDonald's Which one? –  Alex Becker Jul 19 '12 at 2:09
From what I wrote down in the last line, how can you conclude $\lim_{h \to 0} f(a+h) - f(a) = 0$? (i.e. why are you allowed to distribute the limit)? –  WacDonald's Jul 19 '12 at 2:11
@WacDonald's Because $\lim\limits_{h\to 0}\lambda(h)=0$. Thus $$\lim\limits_{h\to 0}(f(a+h)-f(a))=\lim\limits_{h\to 0}(f(a+h)-f(a))-\lim\limits_{h\to 0}\lambda(h)=\lim\limits_{h\to 0}(f(a+h)-f(a)-\lambda(h))$$ –  Alex Becker Jul 19 '12 at 2:20
So, $\lim_{h \to 0} f(a + h) - \lambda(h) = f(a) = \lim_{h \to 0} f(a+h) - \lim_{h \to 0} \lambda(h) $? –  WacDonald's Jul 19 '12 at 2:28

Since $\lambda(h)$ is a linear mapping from $\mathbb{R}^n$ to $\mathbb{R}^m$, it is continuous and $\lambda(0)=0$. Hence $\displaystyle \lim_{h\rightarrow 0}\lambda(h)=\lambda(0)=0$.

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