Show that if $f$ is differentiable at $a$ then one may expand $f$ around $a$ as $$f(x)=f(a)+(x-a)f'(a) +(x-a)E(x)$$ where $E(x) \to 0$ as $x \to a$

If $f$ is differentiable at $a$ then we have $$f'(a)= \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$$

We can multiply both sides by $(x-a)$ and we get $$\begin{align} (x-a)f'(a) & = \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a} (x-a) \\ & = \lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}\lim_{x\to a} (x-a) \end{align}$$

I can't seem to go any further than this, is this a correct way to tackle this problem? A little hint in the right direction would be appreciated and preferred. Thanks very much.


My suggestion would be to define $E(x)$ (for $x \neq a$) as $$ E(x) = \frac{f(x) - f(a)}{x - a} - f'(a),$$ (so that, if you rearrange this, you'll get the equation you wrote down).

It only remains to show that $\lim_{x \to a} E(x) = 0$. For this step, just remind yourself of the definition of the derivative $f'(a)$, and the proof should hopefully be obvious!

  • $\begingroup$ Thanks very much, very simple and concise answer. $\endgroup$ – Patrick Moloney May 13 '17 at 13:34

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