Monday, 6 July 2015

calculus - Limit (without series expansion and l'Hôpital's rule)



limxlnx+x2+1x+x21(lnx+1x1)2=18




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}.


Answer



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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