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Find $$\lim_{x\to 0} \frac{(1+x)^{1/x} - e}{x}$$.

I tried applying L'Hopital's rule but had difficulty with deriving $(1+x)^{1/x}$, and it seemed to me that there is probably a more elegant solution than the horrible derivative WolframAlpha gave (which was also not useful, as the derivative contained $\frac{1}{x}$ as an exponent). Any help would be appreciated.

Limit
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2 Answers2

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First $(u^v)'=vu^{v-1}+u^v v'\ln u$. Next, when applying LHopital's rule the factor that does not become zero can be replaced by its value (this makes application a bit easier). Now

${((1+x)^{1/x}-e)'\over {x'}}= { (1+x)^{1/x} [ {1\over{x(1+x)}}-{{\ln(1+x)}\over{x^2}}}] $. The first factor become $*e*$. The second factor becomes ${x-(1+x)\ln(1+x)}\over{x^2(1+x)}$. Applying Lhopital to this one gives ${-\ln(1+x)}\over{(x+2)(x+1) x}$. As $x\to 0$ the factor of $1\over{(x+2)(x+1)}$ hovers around $*1/2*$ and we need to find $-\ln(1+x)\over x$ which another application of LHopital reduces to $-1\over{1+x}$, which tends to $*-1*$. So final answer is the product of the asterisked quantities, i.e., $-e/2$

Maesumi
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Just apply l'Hôpital. The expression $(1 + x)^{1/x}$ can be written: $$ (1 + x)^{1/x} = \exp\left( \frac{1}{x} \ln (1 + x) \right) $$ That isn't too hard to handle.

vonbrand
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  • For me it kind of is.. If I derive that, won't I still get a derivative with $\frac{1}{x}$? – Limit Feb 01 '13 at 22:31
  • Probably, try it out. BTW, you certainly can apply l'Hôpital several times, until you get continouos numerator and denominator. Or you can cheat and ask e.g. http://WolframAlpha.com... but just to check you are on the right track. – vonbrand Feb 01 '13 at 22:37
  • I've checked WolframAlpha (spoiler: the limit is $-\frac{e}{2}$, but I need to prove the limit formally.. I'm not really sure how to continue from here, I can continue deriving (and yes, I've tried), but the expression $1/x$ remains - I don't understand how to 'get rid' of it. – Limit Feb 01 '13 at 22:44
  • @Limit, absorb the $1 / x$ part into the denominator? – vonbrand Feb 01 '13 at 22:46
  • Differentiate. We get after some simplification $\left(\frac{x-(1+x)\log(1+x)}{x^2\log(1+x)}\right)(1+x)^{1/x}$. The $(1+x)^{1/x}$ part has limit $e$. For the other part, most pleasant to use Taylor series, $\log(1+x)=x-x^2/2+ O(x^3)$. – André Nicolas Feb 01 '13 at 23:28