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Evaluate $$\lim_{x \to 0}\left[\frac{1}{\log (x+ \sqrt{1+x^2})}-\frac{1}{\log (1-x)}\right]$$

Here is my attempt. I know I made some mistake here. I am getting $-\infty$. Please point me to error. Textbook exercise solution is given as $-\dfrac12$. The given limit is $$\lim_{x \to 0}\frac {\ln (1-x)-\ln(x+\sqrt{1+x^2})}{\ln(x+\sqrt{1+x^2})\ln(1-x)}$$ Applying L'Hôpital's rule $$\lim_{x\to0}\dfrac{\dfrac{-1}{1-x}-\dfrac{1}{x+\sqrt{1+x^2}}\left(1+ \dfrac{x}{\sqrt{1+x^2}}\right)}{\dfrac{1}{x+\sqrt{1+x^2}}\left(1+\dfrac{x}{\sqrt{1+x^2}}\right)\ln(1-x)-\dfrac{\ln(x+\sqrt{1+x^2})}{1-x}}$$ is $$\lim_{x\to0}\dfrac{\dfrac{-1}{1-x}-\dfrac{1}{\sqrt{1+x^2}}}{\dfrac{1}{\sqrt{1+x^2}}\ln({1-x})-\dfrac{\ln (x+\sqrt{1+x^2})}{1-x}} = -\infty.$$

Robert Z
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Magneto
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3 Answers3

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There is no need of L'Hopital rule here, this difference is not an indeterminate form! $$\lim_{x \to 0^{+}}\frac{1}{\log (x+ \sqrt{1+x^2})}=+\infty\quad\text{and}\quad \lim_{x \to 0^{+}}\frac{1}{\log (1-x)}=-\infty.$$ Hence $$\lim_{x \to 0^{+}}\left(\frac{1}{\log (x+ \sqrt{1+x^2})}-\frac{1}{\log (1-x)}\right)=+\infty-(-\infty)=+\infty.$$ In a similar way, we have that the limit is $-\infty$ as $x\to 0^-$ (and therefore the limit at $0$ does not exist).

On the other hand, by taking the sum, you may apply Hopital two times (a bit hard) or you may use Taylor expansions: as $x\to 0$, $$\begin{align*}\frac{1}{\log (x+ \sqrt{1+x^2})}+\frac{1}{\log (1-x)}&= \frac{\log (1-x)+\log (1+x+ \frac{x^2}{2}+o(x^2))}{x\cdot (-x)+o(x^2)}\\ &=\frac{(-x-\frac{x^2}{2}+o(x^2))+(x+ \frac{x^2}{2}-\frac{x^2}{2}+o(x^2))}{x\cdot (-x)+o(x^2)}\to \frac{1}{2}.\end{align*}$$

Robert Z
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  • It is ($\infty-\infty$) right? this is indeterminate form. – Magneto Dec 19 '17 at 09:02
  • how is it $+- \infty$ for 1st one and $-+ \infty $ for 2nd one. Whn we put x = 0 we get 1/ log1 = $\infty$. Pls enlighten – Magneto Dec 19 '17 at 09:07
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    @anirudhb NO, it is $+\infty-(-\infty)=+\infty+\infty$ when $x\to 0^+$, and $-\infty-(+\infty)=-\infty-\infty$ when $x\to 0^-$. – Robert Z Dec 19 '17 at 09:19
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    @anirudhb "1/ log1 = ∞" is not enough. You need the sign of this infinity!! – Robert Z Dec 19 '17 at 09:21
  • can u pls elaborate on this? i do not know about this. Atleast point me to some part in textbook. If $x \to 0^+, say x = 0.01, \ln (1-x) = \ln (1-0.01) = \ln 0.99 ~ \ln 1 = 0$ Pls clarify – Magneto Dec 19 '17 at 09:43
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    Note that $\ln(1-x)<0$ for $x\in (0,1)$. Hence, in $x\to 0^+$, $\ln(1-x)$ goes to $0^-$ and $1/0^-=-\infty$. Similar argument for $x<0$. – Robert Z Dec 19 '17 at 10:38
  • Thank you for clarifying. – Magneto Dec 19 '17 at 11:08
  • @anirudhb Any further doubt? – Robert Z Dec 19 '17 at 11:19
  • No Sir, crystall clear. I am really amazwd at the way the people in this forum solve problems. Text books do not provide such clarity. May be it takes time to digest in this way – Magneto Dec 19 '17 at 11:43
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after applying the rules of L'Hopital we get $${\frac {\sqrt {{x}^{2}+1}+1-x}{\ln \left( x+\sqrt {{x}^{2}+1} \right) \sqrt {{x}^{2}+1}+\ln \left( 1-x \right) x-\ln \left( 1-x \right) }} $$ and for $x$ tends to Zero we get that the Limit is undefined

Dr. Sonnhard Graubner
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Let's have $x=\sinh(u)$ with $u\to 0$.

Since $\ln(x+\sqrt{1+x^2})=u\quad$ we get

$f(x)=\dfrac 1u-\dfrac 1{\ln(1-\sinh u)}=\dfrac 1u+\dfrac 1{-\sinh u +o(\sinh u)}=\dfrac 1u-\dfrac 1{-u+o(u)}=\dfrac 2u+o(\frac 1u)$

And this has no limit in zero, since the left and right limits are infinite with different sign and disagree.

zwim
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