# On the Definition of the Derivative

How can I show that if a function $f$ from an open interval of the real numbers to an euclidean space is differentiable at a point $x_0$ in its domain, that is, the limit $\lim_{\epsilon \to 0;\ \epsilon \neq 0}\frac{f(x_0 + \epsilon) - f(x_0)}{\epsilon}$ exists, then the limit $$\lim_{\substack{(x_2, x_1)\to (x_0,x_0);\\ x_1 \leq x_0 \leq x_2;\ x_1 \neq x_2}} \frac{f(x_2)-f(x_1)}{x_2 - x_1}$$ exists and is equal to the derivative of $f$ at $x_0$.

• I struggle to understand the second limit process. Is that a 2D limit $(x_1, x_2) \to (x_0, x_0)$? – mvw Aug 16 '18 at 19:25
• Hint: Set $x_1=x_0$ and set $x_2=x_1+\epsilon$. – The Count Aug 16 '18 at 19:42
• @TheCount Yes, that shows that the limits are equal if the desired limit exists, but it does not show that the desired limit exists. – user109871 Aug 16 '18 at 19:43
• Differentiability is more or less equivalent to the first limit you have written, but I see what you mean. – The Count Aug 16 '18 at 19:46

This is actually rather straightforward. Assume (temporarily) that $x_1 < x_0 < x_2$. Under your assumptions at least one of these inequalities must be strict. Observe $$\frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{f(x_2) - f(x_0)}{x_2 - x_0} \cdot \frac{x_2 - x_0}{x_2 - x_1} + \frac{f(x_1) - f(x_0)}{x_1 - x_0} \cdot \frac{x_0 - x_1}{x_2 - x_1}$$ and $$f'(x_0) = f'(x_0) \cdot \frac{x_2 - x_0}{x_2 - x_1} + f'(x_0) \cdot \frac{x_0 - x_1}{x_2 - x_1}.$$ It follows that $\displaystyle \left| \frac{f(x_2) - f(x_1)}{x_2 - x_1} - f'(x_0) \right|$ is bounded by $$\left| \frac{f(x_2) - f(x_0)}{x_2 - x_0} - f'(x_0) \right| \cdot \frac{x_2 - x_0}{x_2 - x_1} + \left| \frac{f(x_1) - f(x_0)}{x_0 - x_1} - f'(x_0) \right|\cdot \frac{x_0 - x_1}{x_2 - x_1}.$$ Absolute values are not required on the fractions since the terms are positive.
In the general case, if either $x_2 = x_0$ or $x_1 = x_0$ just omit the offending term; the remaining term provides the upper bound.
Now it just needs quantifiers: fix $\epsilon > 0$ and choose $\delta > 0$ so that $0 < |x - x_0| < \delta$ implies $$\left| \frac{f(x) - f(x_0)}{x - x_0} - f'(x_0) \right| < \epsilon.$$
Then $|x_2 - x_1| < \delta$ along with the condition $x_1 \le x_0 \le x_2$ implies simultaneously that $|x_1 - x_0| < \delta$ and $|x_2 - x_0| < \delta$, so that $$\left| \frac{f(x_2) - f(x_1)}{x_2 - x_1} - f'(x_0) \right| < \epsilon \cdot \frac{x_2 - x_0}{x_2 - x_1} + \epsilon \cdot \frac{x_0 - x_1}{x_2 - x_1} = \epsilon.$$
Assume $x_0=0$ and $f'(0)=a$. Then one has $$f(x)=f(0)+ ax + x r(x)$$ whereby $$\lim_{x\to0} r(x)=r(0)=0\ .\tag{1}$$ It follows that $$f(x_2)-f(x_1)=a(x_2-x_1)+x_2 r(x_2)-x_1 r(x_1)$$ and therefore $${f(x_2)-f(x_1)\over x_2-x_1}-a={x_2\over x_2-x_1}r(x_2)-{x_1\over x_2-x_1}r(x_1)\qquad(x_1<x_2)\ .$$ If $x_1\leq0\leq x_2$ we therefore have $$\left|{f(x_2)-f(x_1)\over x_2-x_1}-a\right|\leq \bigl|r(x_2)\bigr|+\bigl|r(x_1)\bigr|\qquad(x_1<x_2)\ ,$$ so that the claim follows from $(1)$.