MHB Derivative of $\frac{1}{x}: \frac{-1}{x^2}$

[Fermat1]

I want to show the derivative of $\frac{1}{x}$ is $\frac{-1}{x^2}$. Well, $\frac{1/(x+h)-1/x}{h}=\frac{-h^2}{x(x+h)}$. Why can't I just plug in $h=0$ to get that the limit is 0?

 

[I like Serena]

I want to show the derivative of $\frac{1}{x}$ is $\frac{-1}{x^2}$. Well, $\frac{1/(x+h)-1/x}{h}=\frac{-h^2}{x(x+h)}$. Why can't I just plug in $h=0$ to get that the limit is 0?

$$\frac{1/(x+h)-1/x}{h}=\frac{-1}{x(x+h)}$$

The value $h=0$ can indeed be plugged in and gives the desired result.

 

[Fermat1]

$$\frac{1/(x+h)-1/x}{h}=\frac{-1}{x(x+h)}$$

The value $h=0$ can indeed be plugged in and gives the desired result.

Thanks. That division is not associative has tripped me up a few times over the years.