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

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.

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This post may give you some ideas. – David Mitra Feb 1 '13 at 23:05

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., $-e/2$

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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.

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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 1 '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. WolframAlpha.com... but just to check you are on the right track. – vonbrand Feb 1 '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 1 '13 at 22:44
@Limit, absorb the $1 / x$ part into the denominator? – vonbrand Feb 1 '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 1 '13 at 23:28