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For the limit

$ \lim_{x \to \infty} (\frac {x+1}{x+2})^x $

could you split it up into the fraction $ \lim_{x \to \infty} (1 - \frac{1}{x+2})^x$

and apply the standard limit $ \lim_{x\to+\infty} \left(1+\frac{k}{x}\right)^x=e^k $

or would you have to apply L'Hopitals rule

edit:

oops, forgot to add a term

Daniel W. Farlow
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Pyrons
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  • See my updated answer. I linked to a PDF from Wolfram|Alpha that outlines the solution you want. – Daniel W. Farlow Feb 14 '15 at 07:18
  • Yes, you can split it, but you could also just write $(x+1)/(x+2) = 1 - (1/(x+2))$, tinker a little with the power, and come to the same conclusion. – user24142 Feb 14 '15 at 07:27

5 Answers5

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The leading term in the denominator of $\frac{x+1}{x+2}$ is $x$. Divide the numerator and denominator by this to get $$ \lim_{x\to\infty}\frac{1+\frac{1}{x}}{1+\frac{2}{x}}. $$ Now, the expressions $\frac{1}{x}$ and $\frac{2}{x}$ both tend to $0$ as $x\to\infty$: $\frac{1}{1}$. Thus, we see that $$ \lim_{x\to\infty}\frac{x+1}{x+2}=\frac{1}{1}=1. $$

Edit: Here's an answer with all of the steps outlined by Wolfram|Alpha (solution to the new limit problem).

Daniel W. Farlow
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4

$$\left( \frac{(x+2) - 1}{x+2} \right)^x = \left( 1 - \frac{1}{x+2} \right)^{x+2} \cdot \left( 1 - \frac{1}{x+2} \right)^{-2} \to e^{-1} \cdot 1 $$

Nicolas Bourbaki
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There's no exponent in your limit. You can split the fraction up just like you have it, but then the limit of 1 goes to 1 and the limit of $\frac 1 {x+2}$ goes to 0, so the two limits exist independantly, thus you can subtract the two limits and get the limit is 1.

Alan
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Hint

Let $$A= \Big(\frac {x+1}{x+2}\Big)^x$$ and take logarithm; so $$\log(A)=x\log\Big(\frac {x+1}{x+2}\Big)$$ Perform the long division and get $$\frac {x+1}{x+2}=1-\frac{1}{x}+\frac{2}{x^2}+\cdots$$ Now, use the fact that, for small $y$, $\log(1+y)\approx y$

I am sure that you can take from here.

Claude Leibovici
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Yes what you did is correct,

Doing L'hopitals requires taking the derivative of $(1+x)^x$ where $a=$ 1 or 2, is really complicated and won't cancel very easily, here is the derivative if you wanted it:

$ \frac d {dx} (1+x)^x = \frac d {dx} e^{xln(1+x)} \\ \ = e^{xln(1+x)} \frac d {dx} (xln(1+x)) \\ \ = e^{xln(1+x)} (ln(1+x)+\frac x {1+x}) \\ \ = (x+1)^x (ln(1+x)+\frac x {1+x}) $

So you can see how complicated it will get. I would go with what you started with, and do this:

$ \lim_{x \to \infty } (\frac {x+1} {x+2}) ^x \\ \ = \lim_{x \to \infty } (1 - \frac {1} {x+2}) ^x \\ \ = \lim_{x \to \infty } (1 + \frac {-1} {x}) ^x (1 - \frac {1} {x}) ^{-2} \\ \ = e^{-1} (1+0)^{-2} \\ \ = e^{-1} $

asosnovsky
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