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Can you prove the derivative of $e^x$ given the definition of $e$, which is $\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x$ without using l'hospital's rule?

e.g. I got to the point where I got $\lim_{x\to0}e^x\left(\dfrac{e^c-1}{c}\right)$, but since our teacher doesn't allow us to use l'hospital's rule or fractional expansion, I'm completely stuck.

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$$\lim_{h\to 0} {e^{x+h}-e^x \over h} = {e^x}\lim_{h\to 0} {e^h-1\over h}$$

Then, set $n = \frac{1}{e^h - 1}$. This means ${1\over n} = e^h - 1$ and thus ${1 \over n} + 1 = e^h$, so $h = \ln\left({1\over n}+1\right)$. It also means as $h \to 0$, $n \to \infty$ so you can rewrite:

$$\lim_{n \to \infty} {{1\over n} \over \ln\left({1\over n}+1\right)} = \lim_{n \to \infty} {1\over n\ln\left({1\over n}+1\right)} = \lim_{n \to \infty} {1\over \ln\left(\color{red}{\left({1\over n}+1\right)^{n}}\right)}$$

Notice that you have the given limit in the denominator:

$$\lim_{n \to \infty}\, \left({1\over n}+1\right)^{n} = e$$

Thus:

$$\lim_{n \to \infty} {1\over \ln\left({1\over n}+1\right)^{n}} = {1\over \ln(e)} = 1$$

Bringing it all together with our original limit:

$$\lim_{h\to 0} {e^h-1\over h} = 1$$

$${e^x}\lim_{h\to 0} {e^h-1\over h} = 1 \cdot e^x = e^x$$

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