0
$\begingroup$

The limit is $$\lim_{x\to0}\left(\frac{\ln(1+x)-\ln(1-x)}{x}\right)$$ Please resolve without using l'hôpital's rule, haven't made it that far yet. Thanks in advance.

$\endgroup$

closed as off-topic by Nosrati, Alexander Gruber Jan 17 at 23:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

5
$\begingroup$

Have you learned how to find limit using definition of derivative yet? if so, you can write it like this:

$\displaystyle \lim_{x \to 0}\dfrac{\ln(1+x) - \ln(1-x)}{x} = \displaystyle \lim_{x \to 0}\left(\dfrac{\ln(1+x)}{x} - \dfrac{\ln(1-x)}{x}\right)= \dfrac{d}{dx}|_{x = 0} \ln(1+x) - \dfrac{d}{dx}|_{x=0} \ln(1-x)=1 - (-1) = 2$

$\endgroup$
  • $\begingroup$ Is this not L'hopitals? $\endgroup$ – John Lou May 24 '17 at 1:27
  • 4
    $\begingroup$ @JohnLou. This is not L'Hopital's Rule. It's applying the definition of derivative to calculate a limit. $\endgroup$ – sharding4 May 24 '17 at 1:28
  • $\begingroup$ No, we haven't even started with derivatives yet. $\endgroup$ – Alvaro Morales May 24 '17 at 1:29
  • $\begingroup$ It seems really similar to L'hopitals, though, as if it were a rudimentary version. $\endgroup$ – John Lou May 24 '17 at 1:30
  • 2
    $\begingroup$ @JohnLou It is not LHR. It bears a superficial resemblance to it, but it's elementary, and LHR isn't, depending as it does on values of derivatives away from the point in question as well as the Cauchy generalized MVT. $\endgroup$ – zhw. May 24 '17 at 22:38
1
$\begingroup$

Let $\displaystyle y=\ln\left(\frac{1+x}{1-x}\right)$. Then

$$e^y=\frac{1+x}{1-x}$$

$$x=\frac{e^y-1}{e^y+1}$$

As $x\to 0$, $y\to0$.

$$\lim_{x\to 0}\frac{\ln(1+x)-\ln(1-x)}{x}=\lim_{y\to0}\frac{y(e^y+1)}{e^y-1}=\lim_{y\to0}\frac{y}{e^y-1}\cdot\lim_{y\to0}(e^y+1)=2$$

$\endgroup$
  • 1
    $\begingroup$ How did you evaluate $$\lim_{y\to0}\frac{y}{e^y-1}$$ without L'Hopital? $\endgroup$ – John Doe May 24 '17 at 1:36
  • $\begingroup$ That's it! Thank you Kwong. $\endgroup$ – Alvaro Morales May 24 '17 at 1:38
  • $\begingroup$ He used lhopitals i guess $\endgroup$ – FickyLicky May 24 '17 at 1:41
  • 1
    $\begingroup$ @Cerkal No. There are many ways to define the constant $e$. Of course, those definitions are equivalent. We can define $e$ as the number such that $\lim_{h\to0}\frac{e^h-1}{h}=1$. en.wikipedia.org/wiki/… $\endgroup$ – CY Aries May 24 '17 at 2:21
1
$\begingroup$

There is a very important and useful Taylor series $$\log \left(\frac{1+x}{1-x}\right)=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}=2 x+\frac{2 x^3}{3}+O\left(x^5\right)$$ which makes $$\frac 1x\log \left(\frac{1+x}{1-x}\right)=2+\frac{2 x^2}{3}+O\left(x^4\right)$$ which shows the limit and also how it is approached.

$\endgroup$
  • $\begingroup$ I still remember the time when I first met this series (20 years ago) and used it to calculate $\log 3$ by putting $x=1/2$. Much better than the usual series for $\log(1+x)$ because it converges faster and has all terms positive. +1 $\endgroup$ – Paramanand Singh May 24 '17 at 18:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.