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$$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$

Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.

Note: WolframAlpha confirms that the result is $\frac{1}{8}$.

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    $\begingroup$ what is $\ln^{-2}{\frac{x+1}{x-1}}$ $\endgroup$ – Aang Sep 5 '12 at 12:01
  • $\begingroup$ That is 1 / {ln[(x+1)/(x-1)]*ln[(x+1)/(x-1)]} . $\endgroup$ – Honore Sep 5 '12 at 12:02
  • $\begingroup$ I haven't found out if I can use this rule or not in Calculus. I think using it is like a misdeed in Mathematics. :-) $\endgroup$ – mrs Sep 5 '12 at 12:36
  • $\begingroup$ I guess you have to know some almost-series facts. Like behavior of $\log(1+t)$ for $t$ near zero, and behavior of $\sqrt{1+t}$ for $t$ near zero. Maybe I'll add more details in a few hours. $\endgroup$ – GEdgar Sep 5 '12 at 12:52
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It is easy to observe that as $x \to \infty$ the argument of both the logarithms tend to $1$ and this is exactly what we need when we are dealing with limit of logarithmic terms. Just add $1$ and subtract $1$ in the argument of log and proceed as follows: \begin{align} L &= \lim_{x \to \infty}\log\frac{x + \sqrt{x^{2} + 1}}{x + \sqrt{x^{2} - 1}}\left(\log\frac{x + 1}{x - 1}\right)^{-2}\notag\\ &= \lim_{t \to 0^{+}}\log\frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}}\left(\log \frac{1 + t}{1 - t}\right)^{-2}\text{ (putting }x = 1/t)\notag\\ &= \lim_{t \to 0^{+}}\log\left(1 + \frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}} - 1\right)\left\{\log\left(1 + \frac{1 + t}{1 - t} - 1\right)\right\}^{-2}\notag\\ &= \lim_{t \to 0^{+}}\log\left(1 + \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\ &= \lim_{t \to 0^{+}}\dfrac{\log\left(1 + \dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)}{\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}}\cdot\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\ &= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\dfrac{\log\left(1 + \dfrac{2t}{1 - t}\right)}{\dfrac{2t}{1 - t}}\cdot\dfrac{2t}{1 - t}\right\}^{-2}\notag\\ &= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\cdot\frac{(1 - t)^{2}}{4t^{2}}\notag\\ &= \frac{1}{8}\lim_{t \to 0^{+}}\frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{t^{2}}\notag\\ &= \frac{1}{8}\lim_{t \to 0^{+}}\frac{2t^{2}}{t^{2}\left\{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}\right\}}\notag\\ &= \frac{1}{8}\notag \end{align} The technique used above is based on the standard limits and is well explained in the comments to this answer.

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  • $\begingroup$ Can you explain how you get second step? $\endgroup$ – Sathasivam K Aug 17 '16 at 15:55
  • $\begingroup$ @SathasivamK: I used substitution $x = 1/t$. I will add this explanation in my answer. $\endgroup$ – Paramanand Singh Aug 17 '16 at 15:56
  • $\begingroup$ It is easy to observe that as x→∞ the argument of both the logarithms tend to 1.how this is possible. I'm self studying so I can't understand .plz explain it $\endgroup$ – Sathasivam K Aug 17 '16 at 15:56
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    $\begingroup$ @SathasivamK: When $x \to \infty$ then $(x + 1)/(x - 1) \to 1$. This is very obvious as $(x + 1)/(x - 1) = (1 + 1/x)/(1 - 1/x)$ and $1/x \to 0$. The first fraction $(x + \sqrt{x^{2} + 1})/(x + \sqrt{x^{2} - 1})$ also tends to $1$ and we can see this by dividing the fraction by $x$ in both numerator and denominator. $\endgroup$ – Paramanand Singh Aug 17 '16 at 15:59
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    $\begingroup$ @SathasivamK: Yes we have $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ as one of the standard limits and this has been used here smartly to get rid of the logarithmic term. That's why I said that argument of log tends to $1$ and thus we add 1 and sub 1 to express it as $\log(1 + x)$ where $x \to 0$ and then use the standard limit related to log. $\endgroup$ – Paramanand Singh Aug 17 '16 at 16:10
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I guess the only way to avoid using series expansion or l'Hôpital's rule is to reduce the limit to known limits. Let $y=1/x$. Then $$ \lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2} = \lim_{y \to 0} \frac{\ln \frac{1+\sqrt{1+y^2}}{1+\sqrt{1-y^2}}}{\left(\ln \frac{1+y}{1-y}\right)^2} = \lim_{y \to 0} \frac{\frac{1}{y^2}\ln\left(\frac{1+\sqrt{1+y^2}}{2}\right) - \frac{1}{y^2}\ln\left(\frac{1+\sqrt{1-y^2}}{2}\right)}{\left(\frac{1}{y} \ln(1+y) - \frac{1}{y} \ln(1-y)\right)^2} $$ Now, using: $$ \lim_{y \to 0} \frac{1}{y} \ln(1\pm y) = \pm 1, \qquad \lim_{y \to 0} \frac{1}{y^2} \ln\left(\frac{1+\sqrt{1 \pm y^2}}{2}\right) = \pm \frac{1}{4} $$ we readily arrive at the result. Limits above can be reduced to one of the classic limits involving the exponential function, namely: $$ \lim_{t \to 0} \frac{\mathrm{e}^t-1}{t} = 1 $$ Indeed: $$ \lim_{y \to 0} \frac{1}{y} \ln(1\pm y) \stackrel{y = \pm \left(\exp(t)-1\right)}{=} \pm \left( \lim_{t \to 0} \frac{t}{\mathrm{e}^t-1}\right) = \pm 1 $$ Using using the substitution $t = \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right)$ we have $$ \lim_{y \to 0} \frac{1}{y^2} \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right) = \lim_{t\to 0} \frac{t}{4 \left(\mathrm{e}^{2t} - \mathrm{e}^t\right)} = \frac{1}{4} \cdot \lim_{t \to 0} \mathrm{e}^{-t} \cdot \lim_{t\to 0} \frac{t}{\left(\mathrm{e}^{t} - 1\right)} = \frac{1}{4} $$ and similarly for the case with $\sqrt{1-y^2}$.

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  • $\begingroup$ Thanks a lot. I do appreciate. Regards $\endgroup$ – Honore Sep 5 '12 at 17:19

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