# Derivative of $e^x:$ What's wrong with this proof?

Note that $$\lim_{h \rightarrow 0} (1 +h)^{1/h} = e$$

Let $f(x) = e^x$. Then $f'(x)$ is given by,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\lim_{h \rightarrow 0} \frac{e^x(e^h-1)}{h} = \lim_{h \rightarrow 0} e^x\cdot \frac{(e^h-1)}{h}$$

Now since $$e = \lim_{h \rightarrow 0} (1 +h)^{1/h},$$ we have $$\lim_{h \rightarrow 0} e^x\cdot \frac{((\lim_{h \rightarrow 0} (1 +h)^{1/h})^h-1)}{h} =e^x\cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (1 +h)^{h/h})-1)}{h} = \\ e^x \cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (1 +h))-1)}{h} = e^x \cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (h))+1-1)}{h} = \\e^x \cdot \lim_{h \rightarrow 0} \frac{\lim_{h \rightarrow 0} h}{h} = e^x \cdot 1 = e^x$$

The problem is you have two different limits and you're using $h$ to represent the variable in both limits. You need to call one of them something besides $h$. Note that this makes the step where you simplify $(1+h)^{h/h} = (1+h)$ invalid.