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I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule.

I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having trouble differentiating $f(x) = (\frac{x+1}{x-1})^{\sqrt{x^2-1}}$. I've used Mathematica, but I won't understand it unless I see the step-by-step process.

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    $\begingroup$ Did you try taking the log of this expression? $\endgroup$ – Simon S Nov 2 '14 at 17:17
  • $\begingroup$ I had not. For some reason I kept trying to apply the chain-rule. $\endgroup$ – Pubbie Nov 2 '14 at 17:18
  • $\begingroup$ You should roughly get the result of $e^2$ without using LHopital. It is much more important to get a feeling. LHoptial is just brute force. $\endgroup$ – Troy Woo Nov 2 '14 at 17:19
  • $\begingroup$ So I managed to take the first derivative, but plugging in $\infty$ doesn't seem to give $e^2$, so I assume I need to take the derivative again? $\endgroup$ – Pubbie Nov 2 '14 at 17:24
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rewrite your term $y=\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}$ in the form $e^{\frac{\ln\left(\frac{(x+1)}{x-1}\right)}{\frac{1}{\sqrt{x^2-1}}}}$
and use L'Hospital.

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  • $\begingroup$ Rewrite what term? Also, plugging in $\infty$ doesn't seem to give $e^2$? Or at least, it isn't obvious that it does. $\endgroup$ – Pubbie Nov 2 '14 at 17:27
  • $\begingroup$ No, plugging infinity doesn't give anything in most real expressions. Yet passing to the limit it does yield the limit $\;e^2\;$ $\endgroup$ – Timbuc Nov 2 '14 at 17:35
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Use that if $\displaystyle\lim_{x\to+\infty}f(x)=L$ then $\displaystyle\lim_{x\to+\infty}\ln(f(x))=\ln(L)$, for $f(x),L>0$.

And that $\ln(a^b)=b\ln(a)$ for $a,b>0$.

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  • $\begingroup$ You're right @Timbuc $\endgroup$ – Surtan Nov 2 '14 at 18:01
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$$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}}=\left(1+\frac2{x-1}\right)^{\sqrt{x^2-1}}=\exp\left(\sqrt{x^2-1}\;\log\left(1+\frac2{x-1}\right)\right)$$

Try now l'Hospital with

$$\lim_{x\to \infty}\frac{\log\left(1+\frac2{x-1}\right)}{\frac1{\sqrt{x^2-1}}}\;,\;\;\text{and then use continuity of the exponential function}$$

Caution: it looks like l'H isn't going to work, but after you do some algebraic order in the resulting slightly messy expression it certainly works.

Added on request:

$$\lim_{x\to \infty}\frac{\log\left(1+\frac2{x-1}\right)}{\frac1{\sqrt{x^2-1}}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac{x-1}{x+1}\left(-\frac2{(x-1)^2}\right)}{-\frac x{(x^2-1)\sqrt{x^2-1}}}=$$

$${}$$

$$=\lim_{x\to\infty}\frac{2\sqrt{x^2-1}}{x}=\lim_{x\to\infty}\frac{2\sqrt{1-\frac1{x^2}}}{1}=2\;\;\implies$$

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

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  • $\begingroup$ I'm still not sure how to evaluate this or how to manipulate it in such a way that the result is $e^2$. $\endgroup$ – Pubbie Nov 2 '14 at 21:01
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Note that: $$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = \left(1 + \frac{2}{x-1}\right)^{\sqrt{x^2-1}} = e^{\sqrt{x^2-1} \log \left(1 + \frac{2}{x-1}\right)}$$

Now we work with the exponent: $$\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \log \left(1 + \frac{2}{x-1}\right) \overset{(1)}{=} \lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \frac{2}{x-1} \overset{(2)}{=} 2$$

Therefore: $$\lim_{x\rightarrow \infty} \left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = e^2$$


Observations.

$(1)$ Note that $\log(1+z) \sim z$ when $z$ goes to zero. Here "$z = \frac{2}{x-1}$".

$(2)$ Clearly $\lim_{x\rightarrow \infty} \frac{\sqrt{x^2-1}}{x-1}= \lim_{x\rightarrow \infty}\sqrt{\frac{(x-1)(x+1)}{(x-1)^2}}= \lim_{x\rightarrow \infty} \sqrt{\frac{x+1}{x-1}}= \lim_{x\rightarrow \infty} \sqrt{1+\frac{2}{x-1}} =1$.

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If you switch variables to $y = x + 1$ you are trying to get this limit: $$\lim_{y \rightarrow \infty} \bigg(1 + {2 \over y}\bigg)^{\sqrt{y^2 + 2y}}$$ $$= \lim_{y \rightarrow \infty} \bigg(1 + {2 \over y}\bigg)^{y\sqrt{1 + {2 \over y}}}$$ Taking logs finishes it pretty quick...

$$\lim_{y \rightarrow \infty} y\sqrt{1 + {2 \over y}}\ln(1 + {2 \over y}) $$ $$= \lim_{y \rightarrow \infty} y\ln(1 + {2 \over y})$$ $$= \lim_{y \rightarrow \infty} {\ln(1 + {2 \over y}) \over {1 \over y}}$$ $$= \lim_{z \rightarrow 0} {\ln(1 + {2 z}) \over {z}}$$ Now apply L'hopital. You could also have applied l'Hopital after the previous step.

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  • $\begingroup$ How exactly does taking a log finish this pretty quick? I'm getting $\infty*0$ as the limit using your method. $\endgroup$ – Pubbie Nov 2 '14 at 23:34
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    $\begingroup$ I wrote it out... $\endgroup$ – Zarrax Nov 3 '14 at 0:53

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