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How would I go about solving this limit ?

$$\lim_{x\to0} \,(1+x)^{\frac{1}{x}}$$

I tried several times to solve it but I can't...

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closed as off-topic by Travis, Davide Giraudo, drhab, Shaun, voldemort Jan 23 '15 at 13:22

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  • 2
    $\begingroup$ Hint: Write $x = \frac{1}{n}$. $\endgroup$ – Travis Jan 23 '15 at 11:31
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    $\begingroup$ think of the limit of $(1+\frac{1}{n})^n$ when $n\rightarrow \infty$. $\endgroup$ – Ofir Schnabel Jan 23 '15 at 11:31
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    $\begingroup$ Rather than saying "I tried several times to solve it but I can't..." which doesn't add anything, you could share what exactly you have tried so far. $\endgroup$ – Hippalectryon Jan 23 '15 at 11:57
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Hint:

Using the binomial theorem with $x=1/n$, you have

$$(1+\frac1n)^n=1+\frac nn+\frac {n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}+\cdots\le1+\frac11+\frac12+\frac1{3!}\cdots.$$ The sum is bounded and every term converges to the corresponding one in the last expression, that we can denote $e$.

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    $\begingroup$ Nice explanation $\endgroup$ – Vim Jan 23 '15 at 12:31
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Hint

Substitute $t = 1/x$ and remember a famous limit

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Hint: rewrite this as $\exp(\frac{\ln(1+x)}{x})$ as apply l'Hopital.

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    $\begingroup$ using l'Hoptial's rule depends on the derivative of exp/log, which in some constructions depends on the value of this very limit. So this could be a circular argument. $\endgroup$ – Matthew Leingang Jan 23 '15 at 11:36

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