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How to solve $$\lim_{x \to \infty}(\dfrac{x}{x+1})^x$$

The answer is $\dfrac{1}{e}$

I can factor the $x$ out to get:

$$\lim_{x \to \infty}\left(\dfrac{x(1)}{x(1+1/x)}\right)^x = \lim_{x \to \infty}\left(\dfrac{1}{1+1/x)}\right)^x$$

How do I further simplify this to get to my limit?

J. M. ain't a mathematician
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K Split X
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2 Answers2

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You almost got it:

$$\left(\frac1{1+\frac1x}\right)^x=\frac1{\left(1+\frac1x\right)^x}\xrightarrow[x\to\infty]{}\frac1e$$

where the limit is gotten using arithmetic of limits...

DonAntonio
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  • Yes I thought about this, but the exponent $x$, why did it not apply to the $1$? – K Split X Mar 19 '17 at 19:24
  • @KSplitX Look more closely: it did. –  Mar 19 '17 at 19:26
  • Yes but then we have $x \to \infty$, and thus we have $1^\infty$ which is an indeterminate form, is it not? – K Split X Mar 19 '17 at 19:27
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    Here you have the number one raised to infinity, which is 1. The problem of the indeterminate form is given when you have "something which is going to one" raised to infinity because you don't know if the number is more or less then one and then you don't know if the power is going to zero or to infinity. – Alberto Andrenucci Mar 19 '17 at 19:29
  • @KSplitX if $a \leq f(x) \leq a$ for all $x$, what can you say about the limit as $x \to \infty$? – Andres Mejia Mar 19 '17 at 19:32
  • @KSplitX An indeterminate of the form $;1^\infty;$ is exactly as Alberto explained to you. In this case you have a fixed, constant number – DonAntonio Mar 19 '17 at 19:33
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    Thanks @AlbertoAndrenucci and Don I understand now – K Split X Mar 19 '17 at 19:34
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This solution does not require L'Hopital. However, it does require the following daunting identity. To learn more about it (trust me, once you've used it once, you'll never want to forget it), see this.

$$\lim\limits_{x \to a}{\phi(x)^{\psi(x)}} = e^{\lim\limits_{x \to a}{[(\phi(x)-1)\psi(x)}]}$$

Using this identity...

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

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Fine Man
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