Why is the derivative not $\lim x\to x_0$ instead of $\lim h\to 0$

Question

As you see in the following image

you can get a tangent when you are moving towards $x_0$ and not $0$.

But if $\lim \to 0$ then you miss $x_0$ because then you are moving at things from $0$ and then you arent working with $x_0$ anymore.

I understood the proof of "derivative as a limit" up until this. I thought that the limit worked on a specific place so in this place it should had been at x0 because that's where you want the tangent to be. And not 0 because the tangent is not there.

I know I am wrong but I dont know why (i know from textbook and from the khanacadamy video) . Could someone explain why you put $\lim x\to 0$ and not $\lim x\to x_0$?

I think I misunderstood limits :(

Thanks beforehand :)

Answer 1

By definition the derivative of $f(x)$ at a point $x=x_0$ is given by (when the limit exists)

$$f'(x_0)=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

now if we define $x-x_0=h\to 0$ the limit becomes

$$f'(x_0)=\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$$

the two definitions are completely equivalent.