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Find the following limit,

$$\lim_{x\to {\infty}} x \ln\left(\frac{x+1}{x-1} \right)$$

I tried this way, that is $$\lim_{x\to {\infty}} x\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}} \right)=\infty \times \ln1= \infty \times 0=0$$

Noticed my logic is horribly wrong. Tested it out on a calculator and the limit should be near $0.8$.

My second try involves $$\lim_{x\to {\infty}} x\ln\left(\frac{x+1}{x-1} \right)=\lim_{x\to {\infty}} x(\ln(x+1)-\ln(x-1))$$

Where I still have no idea how to proceed.

Any hints? Thanks in advance! List them as solutions. I am looking for hints only.

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4 Answers 4

up vote 3 down vote accepted

Hint: Our expression is equal to $$\ln\left(\left(\frac{x+1}{x-1}\right)^x\right).$$

Note that $$\frac{x+1}{x-1}=1+\frac{2}{x-1}.$$

Another approach that will work is to rewrite the expression as $$\frac{\ln\left(\frac{x+1}{x-1}\right)}{\frac{1}{x}},$$ use L'Hospital's Rule once, and then a little algebra.

Remark: Any solution that attempts to calculate "$\infty \cdot 0$" is automatically incorrect.

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Update: Upps, noticed too late, that only hints were wished. I can't make the solution after the first reformulation at the lhs (which were the hint then) invisible, don't know the latex-trick - sorry; somebode else may edit?

If I'm not too dense at the moment I think it is
$$ \begin{eqnarray} \lim_{x \to \infty } \ln\left( \left( 1+{2\over x-1 }\right)^x\right) &=&\lim_{x \to \infty } &\ln\left( \left( 1+{2\over x }\right)^{x+1} \right) \\ &=&\lim_{x \to \infty } &\ln \left( \left( 1+{2\over x }\right)^x \left( 1+{2\over x }\right) \right) \\ &=& &\ln \left( e^2 \left( 1+ 0\right) \right) \\ &=& 2 \end{eqnarray}$$

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You forgot some $\lim$s there. Nice solution nevertheless. –  Pedro Tamaroff Oct 24 '12 at 17:57
    
@Peter: done... –  Gottfried Helms Oct 24 '12 at 18:12
    
Hi gottfried! How did you get from the first step to the. Second step? I have never seen anything like that before, did you add 1 to all the x in the equation? –  Yellow Skies Oct 25 '12 at 0:41
    
And from second step to the third, did you use any series expansion? Looks familar. Thanks! –  Yellow Skies Oct 25 '12 at 0:43
    
@singa: a) yes, when x going to infinity (so no exact value is given for x) and the result is assumed to converge to some value it doesn't matter, whether the limit-variables are increased by 1, or 2 or something finite. b) in the exponent of the parenthese there is now a "+1" - this can be factored out such that the remaining limit-formula equals exactly that known one for the definition of e and $e^x$ –  Gottfried Helms Oct 25 '12 at 1:34

Rewrite it as $$x \ln \left(\dfrac{1+1/x}{1-1/x}\right) = x \left( \ln(1+1/x) - \ln(1-1/x)\right) = x \left( \left( \dfrac1x - \dfrac1{2x^2} + \dfrac1{3x^3} - \cdots\right) - \left( -\dfrac1x - \dfrac1{2x^2} - \dfrac1{3x^3} - \cdots\right)\right) = 2 + \dfrac2{3x^2} + \cdots$$

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Did you use some kind of series? It seems vaguely familiar to me. –  Yellow Skies Oct 24 '12 at 17:58
1  
@SingaporeanDude. This is the Taylor series for $\log(1+y)$ when $\vert y \vert < 1$. $$\log(1+y) = y - \dfrac{y^2}2 + \dfrac{y^3}3 - \dfrac{y^4}4 \pm$$ –  user17762 Oct 24 '12 at 20:32

$0 \cdot \infty$ is not $0$ is an undeterminate form. Write it as fraction, move one of them to the denominator (careful it matters which one).

Another small hint for later: when you need to derivare the $\ln$ of a fraction, what is the $\ln (\frac{a}{b})$?

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