<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] 
Since $$\ln(x+1)= x-\frac{x^2}{2}+O(x^3)~~~and ~~~e^x= 1+x+\frac{x^2}{2}+O(x^3)$$
we get
$$(1+x)^{\frac{1}{x}}= \exp\left(\frac{1}{x}\ln(1+x)\right) = \exp\left(\frac{1}{x}(x-\frac{x^2}{2} +O(x^3))\right) \\=\exp\left(1-\frac{x}{2} +O(x^2)\right) =e\exp\left(-\frac{x}{2}+O(x^2)\right) = e(1- \frac{x}{2}+O(x^2))  $$
Hence $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} = \lim_{x\rightarrow 0} \frac{e(1- \frac{x}{2}+O(x^2))-e}{x} =\lim_{x\rightarrow 0} -\frac{e}{2}+O(x) = \color{blue}{-\frac{e}{2}}$$
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