# Compute $\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$

Compute

$$\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$$

I did

$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x = \lim_{x\to\infty} \exp(x\cdot \ln(\frac{x-2}{x+2})) = \exp( \lim_{x\to\infty} x\cdot \ln(\frac{x-2}{x+2}))$$

But how do I continue? The hint is to use L Hopital's Rule. I tried changing to

$$\exp(\lim_{x\to\infty} \frac{\ln(x-2)-\ln(x+2)}{1/x})$$

This is

$$(\infty - \infty )/0 = 0/0$$

But I find that I can keep differentiating?

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## 6 Answers

A nitpick: $\infty-\infty$ is not 0! It's undefined. Your limit is of the form $0/0$ though.

You can apply L'H'ôpital from the start if you like: $\lim\limits_{x\rightarrow\infty}{x-2\over x+2} =1$, and $\ln 1=0$. So $$\lim_{x\rightarrow\infty} \Bigl(x \ln{x-2\over x+2} \Bigr) =\lim_{x\rightarrow\infty} {\ln{x-2\over x+2}\over1/x} =\lim_{x\rightarrow\infty} {{x+2\over x-2}\cdot{1(x+2)-1(x-2)\over (x+2)^2} \over- 1/x^2 } =\lim_{x\rightarrow\infty} {{-4x^2\over (x+2) (x-2)} }=-4.$$ (use L'Hopital again to evaluate the limit on the right hand side if you like).

So, $$\lim_{x\rightarrow\infty}\Bigl({x-2\over x+2}\Bigr)^x =e^{ \lim\limits_{x\rightarrow\infty}\bigl(x\ln{x-2\over x+2}\bigr)}=e^{-4}.$$

To answer more directly, L'Hôpital applied to $$\lim_{x\rightarrow\infty}{\ln(x-2)-\ln(x+2)\over 1/x}$$ gives you $$\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}.$$ Now simplify: $${{1\over x-2}-{1\over x+2}\over- 1/x^2} =-x^2\Bigl({1\over x-2}-{1\over x+2}\Bigr) = {-4x^2\over (x+2)(x-2)}.$$ So, using L'Hôpital's rule again $$\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2} =\lim_{x\rightarrow\infty} {-4x^2\over (x+2)(x-2)} =\lim_{x\rightarrow\infty} {-8x\over (x+2)+(x-2)} =\lim_{x\rightarrow\infty} {-8x\over2x}=-4.$$

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Hint :

Rewrite limit into form :

$$\lim_{x\to\infty} \left(1+\frac{1}{\left(\frac{x+2}{-4}\right)}\right)^{\left(\frac{x+2}{-4}\right) \cdot \left(\frac{-4x}{x+2}\right)}$$

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This can be done using only the definition of $e$, $$e = \lim_{n\to\infty}(1+1/n)^n.$$ Notice that this implies immediately that $1/e = \lim_{n\to\infty}(1-1/n)^n$ and, more generally, $$\lim_{n\to\infty} (1+ a/n)^n = e^{a n}.$$ We find $$\lim_{x\to\infty} \left(\frac{x-2}{x+2}\right)^x = \lim_{x\to\infty} \left(\frac{1-2/x}{1+2/x}\right)^x = \frac{e^{-2}}{e^2}$$ and so $$\lim_{x\to\infty} \left(\frac{x-2}{x+2}\right)^x = \frac{1}{e^4}.$$

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

$$\lim_{x\to\infty} (1-\frac{4}{x+2})^x = y$$

taking log on both sides we get

$$ln(y) = x ln (1- \frac{4}{x+2})$$

the expansion for $ln (1+r)$ is $r- \frac{r^2}{2} +\frac{r^3}{3}$ .... where r tends to zero

$$ln(y) = x ( \frac{-4}{x+2} - \frac{\frac{-4}{x+2}^2}{2} +\frac{\frac{-4}{x+2}^3}{3} ....)$$

$ln (y) = \frac{-4x}{x+2}$ {rest all terms will terminate to zero}

$$ln (y) =\lim_{x\to\infty} \frac{-4x}{x+2} = -4$$

$$y = \frac{1}{e^4}$$

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you can use $$\left( \frac{x-2}{x+2}\right)^x = \left(1 - \frac{4}{x+2}\right)^x$$ and $(1 + \frac ax)^x \to \exp(a)$,

HTH, AB

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If you want to use LHopital then $\lim_{u\to 0} \frac{\ln(1+u)}{u}=1$ by Lhopital's rule.
$l= \lim_{x\to \infty} (\frac{x-2}{x+2})^x=\lim_{x\to \infty} \exp((x+2)\ln(1-\frac{4}{x+2})-2\ln(1-\frac{4}{x+2}))$
For $u = -\frac{4}{x+2}:$ $l= \lim_{u\to 0}\exp(-4\times\frac{\ln(1+u)}{u}-2\ln(1+u))=\exp(-4)$

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