Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$ Since we have $\frac00$,I Applied L'Hopital rule :
$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$
But as you can see it is getting very ugly.
 A: \begin{aligned}
&\lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}\\
&=\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)}-e}{x}\\
&=e\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)-1}-1}{x}\\
&=e\lim_{x\to 0}\frac{{\frac{1}{x}}\ln(1+x)-1}{x}\\
&=e\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}\\
&=e\lim_{x\to 0}\frac{\frac{1}{1+x}-1}{2x}\\
&=e\lim_{x\to 0}\frac{-1}{2(1+x)}\\
&=\frac{-e}{2}
\end{aligned}
A: Hint
Taylor series of $${(1+x)}^{1/x}=e(1-\frac{1}{2}x+\frac{11}{24}x^2+...)$$
If used you get $-e/2$
A: Without series, only L'Hospital
$$
\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}
$$
we get
$$
\lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\
=\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\
=\lim_{x\to 0}\left(\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}\right)e 
$$
another L'Hospital
$$
= \lim_{x\to 0}\left(\frac{\frac{1}{1+x}+\frac{-x}{(1+x)^2}-\frac{1}{1+x}}{2x}\right)e 
= \lim_{x\to 0}\frac{-1}{2(1+x)^2}e = -\frac{e}{2}\\
$$
A: $$y=\dfrac{(1+x)^{\tfrac1x}-e}{x}$$
$$z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$ Now Taylor
$$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$
$$z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
$$y=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$
