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I am new at computing limits with infinitesimals and I am having trouble solving this one:

$$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}$$

I tried to substitute by means of equivalent infinitesimals and I came up with this:

$$\lim_{x\to0}\frac{x-x}{x^2}$$ But I do not know how to continue. The result must be $-1$. Any help would be appreciated!

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    $\begingroup$ You needed a more accurate estimate of the terms in the numerator, since they cancel to the order you took. $\ln(1+x) \sim 1+x-x^2/2$ and $\ln(1-x) \sim 1-x+x^2/2$ would have done it. $\endgroup$
    – Ian
    Nov 1, 2017 at 13:49
  • $\begingroup$ Are you familiar with the rule of L'Hospital? I wrote an answer using this method. I hope it helps. $\endgroup$
    – Cornman
    Nov 1, 2017 at 13:51
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    $\begingroup$ Easy Just see that: ---------- $$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2} = \lim_{x\to0} \frac{ \ln(1-x^2)}{x^2}=\color{red}{ \lim_{h\to0^+} \frac{ \ln(1-h)}{h}= -1}$$ - $\endgroup$
    – Guy Fsone
    Nov 1, 2017 at 14:58
  • $\begingroup$ What does this have to do with infinitesimals!? You are just using limits! $\endgroup$
    – Zelos Malum
    Nov 1, 2017 at 20:15
  • $\begingroup$ I tried to use equivalent infinitesimals: $\ln(1+x)∼x$ when $x\to0$ and $\ln(1-x)∼-x$ when $x\to0$ @ZelosMalum $\endgroup$
    – Evoked
    Nov 2, 2017 at 17:37

6 Answers 6

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You can't substitute a function with an equivalent if sums are involved. Here it's better to use Taylor expansion: $$ \ln(1+x)=x-\frac{x^2}{2}+o(x^2) $$ so your limit becomes $$ \lim_{x\to0}\frac{(x-x^2/2)+(-x-x^2/2)+o(x^2)}{x^2} $$

As you see, $x$ and $-x$ cancel out, but there's something of the order of $x^2$ remaining.

Comment

You could use equivalents by noticing that $\ln(1+x)+\ln(1-x)=\ln(1-x^2)$, which is equivalent to $-x^2$, but this works in the particular case and would not help in a case such as $$ \lim_{x\to0}\frac{x-\ln(1+x)}{x^2} $$

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Use

  • $\ln(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3} - \cdots$

  • $\ln(1-x) = -x -\frac{x^2}{2}-\frac{x^3}{3}-\cdots$

Then you have $$\ln(1+x)+\ln(1-x) = -x^{2} -\frac{x^4}{2} + \cdots =x^{2}\cdot \left(-1 -\frac{x^2}{2} + \cdots\right)$$

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  • $\begingroup$ The last expression Inside the parentheses must start with $-1$. $\endgroup$
    – zipirovich
    Nov 1, 2017 at 13:57
  • $\begingroup$ @zipirovich Thanks. Will edit :) $\endgroup$
    – C.S.
    Nov 1, 2017 at 14:31
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$$\begin{align}\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2} &= \lim_{x\to0} \frac{-\ln(1-x^2)}{-x^2} = -1 \end{align}$$

Since $\lim_{x\to 0} \dfrac{\ln(x + 1)}{x} = 1$.

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We apply L'Hospital:

$f(x)=\ln(x+1)+\ln(1-x)$

$g(x)=x^2$

Note that $f(0)=g(0)=0$.

It is $f'(x)=\frac{1}{x+1}+\frac{1}{x-1}$ and $g'(x)=2x$. Note that $f'(0)=g'(0)=0$ so we apply L'Hospital a second time:

$f'(x)=-\frac{1}{(x+1)^2}-\frac{1}{(1-x)^2}$ and $g''(x)=2$

With $f''(0)=-2$ and $g''(0)=2$. Hence:

$\lim_{x\to 0} \frac{\ln(x+1)+\ln(1-x)}{x^2}=\frac{-2}{2}=-1$

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    $\begingroup$ It should be $g(x)=x^2$. Double l’Hôpital can be avoided by rewriting $f'(x)=\frac{2x}{x^2-1}$, so the limit becomes $\lim_{x\to0}\frac{1}{x^2-1}$ after simplifying. $\endgroup$
    – egreg
    Nov 1, 2017 at 14:01
  • $\begingroup$ Thank you. That was a dumb typo I edited it. I like your rewriting to avoid a 2nd time L'Hospital! $\endgroup$
    – Cornman
    Nov 1, 2017 at 14:17
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By L'Hopital Rule we have $$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2} = \lim_{x\to0} \frac{\frac{1}{1+x}-\frac{1}{1-x}}{2x} =\lim_{x\to0} \frac{\frac{-2x}{1-x^2}}{2x} =\lim_{x\to0} \frac{-1}{1-x^2}= -1$$

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$$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}=\lim_{x\to0} \ln(1-x^2)^{1/x^2}=\lim_{t\to\infty} -\ln(1-\dfrac1t)^{-t}=-1$$

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