I was trying to solve the derivative of $e^{x}$ the traditional way with the definition of the derivative: $$ \lim_{h\rightarrow 0}\frac{e^{x+h}-e^{x}}{h} $$ so I solved like this:
$$\lim_{h\rightarrow 0}\frac{e^{x+h}-e^{x}}{h} = \lim_{h\rightarrow 0}\frac{e^{x}\cdot e^{h} -e^{x}}{h}= e^{x}\cdot \lim_{h\rightarrow 0}\frac{e^{h} -1}{h}=e^{x}$$ where I solved $\lim_{h\rightarrow 0}\frac{e^{h} -1}{h}$ numerically to get 1 (which was expected because I've done this with implicit differentiation and with Taylor Series as proofs for the derivative of $e^{x}$ to be itself).
So how to solve $\lim_{h\rightarrow 0}\frac{e^{h} -1}{h}$ non-numerically and without L'Hôspital's rule?
Let $t = e^h -1$. Then we obtain that $h = \ln(1+t)$. Observe that $h\rightarrow 0$ implies that $t\rightarrow 0$ and we have:
$$\lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}= \lim_{h\rightarrow 0}\frac{e^x(e^h-1)}{h} = \lim_{t\rightarrow 0}e^x\cdot \frac{t}{\ln (1+t)}=e^x\cdot \lim_{t\rightarrow 0}\frac{1}{\frac{1}{t}\ln (1+t)}=$$ $$= e^x\cdot \frac{1}{\displaystyle \lim_{t\rightarrow 0} \ln (1+t)^{\frac{1}{t}}}= e^x\cdot \frac{1}{\ln \displaystyle \lim_{t\rightarrow 0} (1+t)^{\frac{1}{t}}}=e^x\cdot\frac{1}{\ln e} =e^x.$$
In this calculus we use the fundamental limit: $\displaystyle \lim_{t\rightarrow 0}(1+t)^{\frac{1}{t}} =e$.