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I got this problem in my math.

Evaluate $$\lim_{x\rightarrow 0} {\frac{(1+x)^{\frac{1}{x}}-e}{x}}$$

I tried applying the L'Hospital rule, to get $$\lim_{x\rightarrow 0} (1+x)^{\frac{1}{x}}\frac{1}{x^2}(\frac{x}{1+x}-ln(1+x))$$ I don't know how to proceed after this.

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  • $\begingroup$ I think this might be a plain derivative in disguise.. $\endgroup$ – Cameron Williams Nov 17 '13 at 6:16
  • $\begingroup$ @CameronWilliams: Indeed $\endgroup$ – meta_warrior Nov 17 '13 at 6:30
  • $\begingroup$ I think that the derivative of the numerator is wrong. Please check. In any manner, what freak_warrior suggested is the good way to go. $\endgroup$ – Claude Leibovici Nov 17 '13 at 6:31
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Hint: Use $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=\lim_{x\rightarrow 0} \frac{e^{\ln{(1+x)^{\frac{1}{x}}}}-e}{x}=\lim_{x\rightarrow 0} \frac{e^{\frac{\ln{(1+x)}}{x}}-e}{x}$$

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  • $\begingroup$ Nice solution freak_warrior......... $\endgroup$ – juantheron Nov 17 '13 at 6:16
  • $\begingroup$ What do I do after this? $\endgroup$ – GTX OC Nov 17 '13 at 7:07
  • $\begingroup$ @GTXOC: Use L'Hopital rule. $\endgroup$ – meta_warrior Nov 17 '13 at 7:09
  • $\begingroup$ Can you give me the full solution? I am still not getting it. $\endgroup$ – GTX OC Nov 17 '13 at 7:37
  • $\begingroup$ Hint: Can you prove that $\lim_{x\to0}\frac{\ln{(1+x)}}{x}=1$ @GTXOC $\endgroup$ – meta_warrior Nov 17 '13 at 7:40
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You can proceed as follows $$\lim_{x\to 0}(1+x)^{1/x}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$ $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$

apply L'Hospital's rule for $\frac00$ form of second limit as follows $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{d}{dx}\frac{x}{1+x}-\frac{d}{dx}\ln(1+x)}{\frac{d}{dx}x^2}$$

$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{1}{(1+x)^2}-\frac{1}{(1+x)}}{2x}$$ $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{-x}{2x(x+1)^2}$$ $$=-\frac12\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{1}{(x+1)^2}$$ $$=-\frac12(e)\cdot (1)$$ $$=-\frac e2$$

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