Geometrical interpretation of $\lim_{h \to 0} \frac{f(x_0+h) - f(x_0+h) }{2h}$ Let $f:D \subset \Bbb R\to \Bbb R$ is $f$ differentiable at $x_0$ Is it true that:
$$f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h) - f(x_0+h) }{2h}$$
$$f'(x_0)=\lim_{h \to 0} \frac{f(x_0+2h) - f(x_0)}{h} \text{ ?}$$
Write the geometrical interpretation of each case.
For the first one I have already proved that are equals and the reciprocal is not true, but I do not know what is the geometrical interpretation.
Any ideas?
 A: \begin{align}
& \lim_{h \to 0} \frac{f(x_0+2h) - f(x_0)} h \\[10pt]
= {} & 2\lim_{h \to 0} \frac{f(x_0+2h) - f(x_0)}{2h} \\[10pt]
= {} & 2\lim_{k\to\text{what?}} \frac{f(x_0+k) - f(x_0)} k \qquad \text{where } k = 2h \\[15pt]
& \text{But as }h\to0\text{ then }k=2h\to 0, \text{ so we have:} \\[10pt]
= {} & 2\lim_{k\to0} \frac{f(x_0+k) - f(x_0)} k \\[10pt]
= {} & 2f'(x_0).
\end{align}
For the first one, you wrote:
$$
f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h) - f(x_0+h) }{2h}
$$
If that is what you meant, then the limit is always $0$ because the numerator is something minus itself, so it's not $f'(x_0)$ except when that is $0$.  However, if you had
$$
f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h) - f(x_0-h) }{2h}
$$
then that is correct.  The geometric interpretation is that $\dfrac{f(x_0+h) - f(x_0-h) }{2h}$ is the slope of the line through two points on the graph close to $(x_0,f(x_0)$, and as those points approach $x_0$, that slope approaches the slope of the curve at $(x_0,f(x_0)$.
A: 
As $h$ goes to $0$ the line between $(x_0-h,f(x_0-h))$ and $(x_0+h,f(x_0+h))$ gets closer to the tangent of the graph of $f$ at $x=x_0$.
