Evaluating $\lim_{x\to 0}\frac{\ln(1+x)-\ln(1-x)}{\arctan(1+x)-\arctan(1-x)}$ without differentiation

Question

$$\lim_{x\to 0}\frac{\ln(1+x)-\ln(1-x)}{\arctan(1+x)-\arctan(1-x)}$$

So, I have this limit and I'm trying to solve this limit without differentiation.

I tried some steps, but they didn't come out well, and now I have no idea how to solve this. I know that the limit of $$\lim_{x\to 0}\frac{\arctan x }{x} = 1$$ but how is that going to help me in this case?

Answer 1

Hint:

$$\arctan p+\arctan q=\arctan\Bigl(\frac{p+q}{1-pq}\Bigr)\quad\text{if }pq<1. $$

Also, $\;\ln(1+x)=x+o(x)$, $\;\ln(1-x)=-x+o(x)$.

Answer 2

You can use the known limit $$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1 $$ Then you can rewrite your limit as $$ \lim_{x\to0}\left(\frac{\ln(1+x)}{x}-\frac{\ln(1-x)}{x}\right) \frac{x}{\arctan(1+x)-\arctan(1-x)} $$ Since the part in parentheses has limit $2$, you can just compute the limit of the remaining fraction; better, of its reciprocal. With a similar trick $$ \lim_{x\to0}\frac{\arctan(1+x)-\pi/4}{x}=\lim_{y\to\pi/4}\frac{y-\pi/4}{\tan y-1}= \lim_{z\to0}\frac{z}{\tan(z+\pi/4)-1}=\lim_{z\to0}\frac{z}{\tan z}\frac{1-\tan z}{2}=\frac{1}{2} $$ with the substitutions $y=\arctan(1+x)$ and $y=z+\pi/4$.

Can you finish?

Answer 3

We will need the result $$\lim_{x\to0}\frac{\ln(1+x)}{x} = 1$$

With this, and the result given about the arctan limit, we can get our answer.

First, we have - \begin{align} \arctan(1+x)-\arctan(1-x)&=\arctan\left(\frac{(1+x)-(1-x)}{1+(1+x)(1-x)}\right)\\ &=\arctan\left(\frac{2x}{2-x^2}\right) \end{align}

So, \begin{align} \lim_{x\to 0}\frac{\ln(1+x)-\ln(1-x)}{\arctan(1+x)-\arctan(1-x)} &= \lim_{x\to 0}\frac{\ln\left(1+\frac{2x}{1-x}\right)}{\arctan\left(\frac{2x}{2-x^2}\right)}\\ &=\lim_{x\to 0}\left(\frac{\ln\left(1+\frac{2x}{1-x}\right)}{\frac{2x}{1-x}}\right)\lim_{x\to 0}\left(\frac{\frac{2x}{2-x^2}}{\arctan\left(\frac{2x}{2-x^2}\right)}\right)\lim_{x\to 0}\frac{\frac{2x}{1-x}}{\frac{2x}{2-x^2}}\\ &=1.1.\lim_{x\to 0}\frac{2-x^2}{1-x}\\ &=2 \end{align}

Answer 4

$\lim_{u\to 0}\frac {u}{\tan u}=1.$ So for $|u|<\pi/2$ let $u=\arctan v.$ Then $1=\lim_{u\to 0}\frac { \arctan v}{v}=\lim_{v\to 0}\frac {\arctan v}{v}.$

So $\arctan v=vF(v)$ where $\lim_{v\to 0}F(v)=1.$

From the angle-sum formulas of trigonometry, when $x^2\ne 2$ we have $$\arctan (1+x)-\arctan (1-x)=\arctan \frac {(1+x)-(1-x)}{1+(1+x)(1-x)}=$$ $$=\arctan \frac {2x}{2-x^2}=$$ $$=\frac {2x}{2-x^2}F(\frac {2x}{2-x^2}).$$

There are many ways to show that $\log (1+x)=xG(x)$ when $|x|<1$ where $\lim_{x\to 0}G(x)=1.$

For example for $x\ge 0$ we have $$x= \int_1^{1+x}(1)dt\ge \int_1^{1+x}(1/t)dt=\log (1+x)\ge \int_1^{1+x}(1+x)^{-1}dt=x(1+x)^{-1}$$ and there is a similar calculation for $-1<x<0.$

So when $|x|<1$ we have $\log (1+x)-\log (1-x)=xG(x)-(-xG(-x))=x(G(x)+G(-x)).$

Answer 5

All this needs is as $c \to 0$, $\dfrac{\ln(1+c)}{c} \to 1$ and $\dfrac{\arctan(c)}{c} \to 1$.

Since $\ln(a)-\ln(b) =\ln(a/b) $ and $\arctan(a)-\arctan(b) =\arctan(\dfrac{a-b}{1+ab}) $,

$\begin{array}\\ f(x) &=\dfrac{\ln(1+x)-\ln(1-x)}{\arctan(1+x)-\arctan(1-x)}\\ &=\dfrac{\ln(\frac{1+x}{1-x})}{\arctan(\frac{2x}{1+(1+x)(1-x)})}\\ &=\dfrac{\ln(\frac{1+x}{1-x})}{\arctan(\frac{2x}{2-x^2})}\\ \end{array} $

Let $\dfrac{1+x}{1-x} =1+c $. Then $1+x = (1+c)(1-x) =(1+c)-x(1+c) $ so $x(2+c) = c $ or $x =\dfrac{c}{2+c} $.

Also,

$\begin{array}\\ \dfrac{2x}{2-x^2} &=\dfrac{2\dfrac{c}{2+c}}{2-(\dfrac{c}{2+c})^2}\\ &=\dfrac{2c(2+c)}{2(2+c)^2-c^2}\\ &=\dfrac{2c(2+c)}{2(4+4c+c^2)-c^2}\\ &=\dfrac{2c(2+c)}{8+8c+c^2}\\ \end{array} $

Then

$\begin{array}\\ f(x) &=f(\dfrac{c}{2+c})\\ &=\dfrac{\ln(1+c)}{\arctan(\frac{2c(2+c)}{8+8c+c^2})}\\ &=\dfrac{\ln(1+c)}{\arctan(c\frac{4+2c}{8+8c+c^2})}\\ \end{array} $

As $c \to 0$, $\dfrac{\ln(1+c)}{c} \to 1$ and $\dfrac{\arctan(c\frac{4+2c}{8+8c+c^2})}{c} \to \dfrac{\arctan(\frac{c}{2})}{c} \to \dfrac12 $ so $f(x) \to 2 $.