How to find the limit: $$\lim_{x \to 0} \frac{e^{(1+x)^{1/x}}-(1+x)^{\frac{e}{x}}}{x^2}$$ I've tried the L'Hôpital's Rule, but failed. Any idea? Thanks in advance!

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We can use L'Hôpital's rule. Let $$f(x)=(1+x)^{\frac{1}{x}}$$ and $$\lim_{x \to 0}f(x)=e$$

Also $$f'(x)=f(x)\:\frac{x-ln(1+x)}{x^2(1+x)}$$ so

$$\lim_{x\to 0}f'(x)=e \times \lim_{x \to 0}\frac{x-ln(1+x)}{x^2(1+x)}=\frac{e}{2}$$

Now $$L=\lim_{x \to 0}\frac{e^{f(x)}-f(x)^e}{x^2}$$ Using L'Hôpital's Rule

$$L=\lim_{x \to 0}f'(x) \times \lim_{x \to 0}\frac{e^{f(x)}-ef(x)^{e-1}}{2x}=\frac{e}{2} \times \lim_{x \to 0}\frac{e^{f(x)}-ef(x)^{e-1}}{2x}$$ $\implies$

$$L=\frac{e^2}{4}\times \lim_{x \to 0}\frac{e^{f(x)-1}-f(x)^{e-1}}{x}$$ Again applying L'Hôpital's Rule

$$L=\frac{e^2}{4} \times \lim_{x \to 0}f'(x) \times \lim_{x \to 0}\frac{e^{f(x)-1}-(e-1)f(x)^{e-2}}{1}$$ $\implies$

$$L=\frac{e^2}{4} \times \frac{e}{2} \times \left(e^{e-1}-(e-1)e^{e-2}\right)=\frac{e^{e+1}}{8}$$

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    $\begingroup$ Very good answer. +1 $\endgroup$ – Paramanand Singh Nov 18 '15 at 4:02
  • $\begingroup$ Beautiful answer, thanks. $\endgroup$ – easymath3 Nov 18 '15 at 4:06

Do the Taylor Expansion around $x=0$, then you get: $$ e^{(1+x)^{1/x}}-(1+x)^{\frac{e}{x}}=\frac{1}{8} e^{1+e} x^2-\frac{1}{16} e^{1+e} (3+e) x^3+\mathcal{O}(x^4) $$ Thus the answer is $\displaystyle\frac{1}{8}e^{1+e}$.

  • $\begingroup$ Thanks, I think Taylor Expansion is too hard and brute force if without computer, any other way? $\endgroup$ – easymath3 Nov 18 '15 at 2:20

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