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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

Martin Sleziak
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1 Answers1

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The hint:

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

Michael Rozenberg
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  • Hi, I think the first equality is supposed to be true because $2x\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)+\frac{ex^2}{2}}\to 1$. But it seems to me that $x\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)+\frac{ex^2}{2}}\to 1$. So I request your help in understanding what I am missing in understanding why there should be a 2 in the denominator after the first equality. Thanks. – Koro Jan 18 '22 at 21:45