Define the natural number e by e=lim.
Then, I can prove \lim_{x\to 0} \frac{e^x-1}{x}=1.
Let z=e^x-1. Then, x=\ln(z+1) and \lim_{x\to 0} \frac{e^x-1}{x}=\lim_{z\to 0} \frac{z}{\ln(z+1)}=\frac{1}{\ln e}=1\text{.}
Using a similar trick (without L'Hoptial's rule), can we prove \lim_{x \to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}?
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