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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?

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marked as duplicate by Guy Fsone, J. M. is a poor mathematician, Claude Leibovici calculus Nov 12 '17 at 6:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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...

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  • $\begingroup$ Yes I thought about this, but the exponent $x$, why did it not apply to the $1$? $\endgroup$ – K Split X Mar 19 '17 at 19:24
  • $\begingroup$ @KSplitX Look more closely: it did. $\endgroup$ – user228113 Mar 19 '17 at 19:26
  • $\begingroup$ Yes but then we have $x \to \infty$, and thus we have $1^\infty$ which is an indeterminate form, is it not? $\endgroup$ – K Split X Mar 19 '17 at 19:27
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    $\begingroup$ 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. $\endgroup$ – Alberto Andrenucci Mar 19 '17 at 19:29
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    $\begingroup$ Thanks @AlbertoAndrenucci and Don I understand now $\endgroup$ – 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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    $\begingroup$ I think you left out some quite important conditions on the limits of $\;\phi,\,\psi\;$ ... $\endgroup$ – DonAntonio Mar 20 '17 at 10:20

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