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I asked recently in this question how to use the definition of

$$e:= \lim_{x\to\infty}\left( 1+\frac1x \right)^x$$

to show that

$$\lim_{x\to -\infty}\left( 1+\frac1x \right)^x = e.$$

A helpful answer said that

$$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$

and $-(\eta + 1) \xrightarrow[]{\eta \to -\infty} \infty$.

It struck me as a brilliant way to solve the problem, and I wondered how one might come up with it. Is the rearrangement done in this answer an instance of a more general strategy, or is it more or less just fiddling until one gets the right form?

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Eric Auld
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2 Answers2

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Hint : Given limit is of the form $1^\infty$, It's a indeterminate form. You can have a look of this to find the limit in such cases.

Why is $1^{\infty}$ considered to be an indeterminate form

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Shravan40
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  • I'm not sure this is relevant, since it is an explanation of indeterminate forms, and recommends taking the logarithm, which is not how the above problem was solved. Thanks anyway – Eric Auld Oct 14 '13 at 17:41
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In general, if $h(x)=1+O(x^{-2})$ then $\lim_{x\to\infty} h(x)^x = 1$.

Now, let $h(x)=\left(1-\frac{1}{x^2}\right)=\left(1-\frac{1}{x}\right)\left(1+\frac 1x\right)$.

Since $h(x)^x\to 1$ as $x\to\infty$ and $(1+1/x)^x\to e$ as $x\to\infty$ we have that $\left(1-\frac 1x\right)^x\to \frac{1}{e}$ as $x\to\infty$.

But then $\left(1-\frac1x\right)^{-x}\to e$ as $x\to\infty$. Replacing $x$ with $-x$, we see that:

$$\lim_{x\to -\infty} \left(1+\frac{1}{x}\right)^x=e$$

So all you have to prove is the first claim.

Thomas Andrews
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