I need help solving a limit, without using l'hôpital's rule.

Question

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.

Answer 1

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

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

Answer 2

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

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

Answer 3

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.