Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

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?

share|cite|improve this question
up vote 3 down vote accepted

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 $$ \lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2} =\lim_{x\rightarrow\infty} {-4x^2\over (x+2)(x-2)} =\lim_{x\rightarrow\infty} {-8x\over (x+2)+(x-2)} =\lim_{x\rightarrow\infty} {-8x\over2x}=-4. $$

share|cite|improve this answer

Hint :

Rewrite limit into form :

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

share|cite|improve this answer

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

share|cite|improve this answer

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

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

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}

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

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

share|cite|improve this answer

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


share|cite|improve this answer

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}: $ $l= \lim_{u\to 0}\exp(-4\times\frac{\ln(1+u)}{u}-2\ln(1+u))=\exp(-4)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.