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

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?

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  • $\begingroup$ Are you familiar with Taylor series? $\endgroup$ – Monadologie May 18 '19 at 21:08
  • $\begingroup$ @Monadologie What if you solve it without the Taylor series? Is that possible? Just by using limit properties? $\endgroup$ – wolly May 18 '19 at 21:09
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    $\begingroup$ @Monadologie If the OP doesn't want differentiation, I greatly doubt he wants Taylor series... $\endgroup$ – DonAntonio May 18 '19 at 21:09
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    $\begingroup$ @wolly: You write that you've "tried some steps, but they didn't come out well". Please include those steps, even if they're wrong. This can help others avoid wasting time (theirs and yours) duplicating your effort and/or telling you things you already know; also, someone may be able to identify a simple flaw in your strategy (sign error or something). $\endgroup$ – Blue May 18 '19 at 21:17
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    $\begingroup$ @Blue Fine but you won't like what you will see(ibb.co/JQrpdL1) $\endgroup$ – wolly May 18 '19 at 21:26
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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}

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Hint:

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

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

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  • $\begingroup$ How are ln(1+x)=x and ln(1-x)=-x? I know that ln(xy)=lnx+lny and ln(x/y)=lnx-lny. Is that what you used? $\endgroup$ – wolly May 18 '19 at 21:29
  • $\begingroup$ It's not equal to $x$ (or $-x$), but to $x$ with a complementary term very small w.r.t. $x$ (generically denoted $o(x)$). This is usually denoted as $\ln(1+x)\sim_{x\to 0} x$, in asymptotic analysis. $\endgroup$ – Bernard May 18 '19 at 21:34
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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?

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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 $.

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$\lim_{u\to 0}\frac {u}{\tan u}=1.$ 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 $\log (1+x)-\log (1-x)=xG(x)-(-xG(-x))=x(G(x)+G(-x)).$

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  • $\begingroup$ $\cos (A-B)=\cos A \cos B+\sin A\sin B$ and $\sin (A-B)=\sin A\cos B-\sin B\cos A.$ So when none of $\cos A, \cos B, \cos (A-B)$ is $0$ we have $\tan (A-B)=(\tan A-\tan B)/(1+\tan A \tan B).$ So with $A=\arctan u$ and $B=\arctan v$ we have $\tan (\arctan u-\arctan v)=(u-v)/(1+uv).$ So $$\arctan u -\arctan v=\arctan ((u-v)/(1+uv)).$$ $\endgroup$ – DanielWainfleet May 18 '19 at 22:39

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