Finding $ \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}$ without using series expansion.

Question

Finding $\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}$ without using series expansion.

I am trying to solve it using D L Hopital Rule

So $\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}\cdot \frac{1}{x^2}\bigg[\frac{x}{1+x}-\ln(1+x)\bigg]+\frac{e}{2}}{2x}$

again i want to Diff with r to $x,$ but this is to complex

please help me how to short my calculation , Thanks

Answer 1

The hint:

$$\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)+\frac{ex^2}{2}}{2x^3}=$$ $$=\frac{1}{2}\lim_{x\rightarrow0}\frac{\left((1+x)^{\frac{1}{x}}-e\right)\left(\frac{x}{1+x}-\ln(1+x)\right)+e\left(\frac{x}{1+x}-\ln(1+x)\right)+\frac{ex^2}{2}}{x^3}=$$ $$=\frac{1}{2}\left(\left(-\frac{e}{2}\right)\cdot\left(-\frac{1}{2}\right)+\frac{2}{3}e\right)=\frac{11e}{24}.$$