<s>[INST] How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$? 
How can I calculate this limit?
  $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$  

I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
 [/INST] By inequality
$$x-\frac{x^2}{2}\leq \log (1+x) \leq x-\frac{x^2}{2}+\frac{x^3}{3}$$
$$1-\frac{x}{2}\leq \frac{\log (1+x)}{x} \leq 1-\frac{x}{2}+\frac{x^2}{3}$$
we have that
$$e^{1-\frac{x}{2}}\leq (1+x)^\frac{1}{x}=e^{\frac{\log (1+x)}{x}} \leq e^{1-\frac{x}{2}+\frac{x^2}{3}} $$
thus
$$-\frac{e}2\le\frac{e^{1-\frac{x}{2}}-e}{x}\leq \frac{(1+x)^{\frac{1}{x}}-e}{x} \leq \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x}\to -\frac{e}2$$
indeed
$$ \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x} =e \cdot \frac{ e^{-\frac{x}{2}+\frac{x^2}{3}}-1}{ -\frac{x}{2}+\frac{x^2}{3} }\cdot\frac{-\frac{x}{2}+\frac{x^2}{3}}{x}\to e\cdot1\cdot -\frac12=-\frac{e}2$$
therefore for squeeze theorem
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=-\frac{e}2$$
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