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I'm trying to show that $$\lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} = -\frac{e}{2}$$

At first it seemed like a routine application of L'Hospital's rule, but my standard bag of tricks isn't working. The $e$ in the numerator prevents any log trickery from separating nicely, and the limit being negative seems to also preclude analyzing the limit of the log.

I tried to interpret this as the derivative of some function at a point, say, $g(u) = u^{\frac{1}{u-1}}$ and the point $u = 1$, but evaluating $g'(1)$ just got worse, and I had concerns about differentiability of $g$ there. Would choosing a different function work out better?

I tried fiddling with one of the limit definitions for $e$ because the first term in the numerator tends to $e$ as $x\to 0$, but the function we're taking the limit of is not continuous at $x=0$ and so moving the limit in was a no-go.

Edit: the $e$ in the numerator seems critical, as the limit diverges without it.

I have a feeling I'm missing something simple. Any hints/solutions would be appreciated.

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Hint: can you see about changing the numerator using a simple substitution so it looks like the typical limit that gives you e? –  Amzoti Nov 28 '12 at 3:55

3 Answers 3

up vote 3 down vote accepted


$$(1+x)^{1/x}=e^{\frac{1}{x}\log(1+x)}\xrightarrow [x\to 0]{}e^1=e$$

and thus we can apply L'H:


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Wow, I guess I just gave up too early when I tried this method. Thanks, I got it! –  sourisse Nov 28 '12 at 4:04


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

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I think I needed a bit more of a hint than this, but thanks anyhow. –  sourisse Nov 28 '12 at 4:07
@sourisse: You are welcome. –  Mhenni Benghorbal Nov 28 '12 at 4:45

Using Taylor polynomials for $x$ small, $$ \frac{(1+x)^{1/x}-e}x=\frac{e^{\frac1x\,\log(1+x)}-e}x=\frac{e^{\frac1x\,(x-x^2/2+O(x^3)))}-e}x=\frac{e^{1-x/2+O(x^2)}-e}x=e\,\frac{e^{-x/2+O(x^2)}-1 }x=e\,\frac{-\frac x2+O(x^2)}x =e\,\left(-\frac12+O(x)\right)\to-\frac e2. $$

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