If $\left|\lim_{x\to\frac\pi2^-}(1+\tan x)\big\{(1+\tan x)\ln\left(\dfrac{1+\tan x}{2+\tan x}\right)+1\big\}\right|$ is $L$, find the value of $4L$.

Question

If $\left|\lim_{x\to\frac\pi2^-}(1+\tan x)\big\{(1+\tan x)\ln\left(\dfrac{1+\tan x}{2+\tan x}\right)+1\big\}\right|$ is $L$, find the value of $4L$.

To work out the approach, I wrote it informally as $(1+\tan x)(P+1)$

I see that $P$ is $\infty\times0$. So, I wrote it as $\frac00$ form and then applied L'Hopital rule. And got $P=-1$

So, the overall expression is also $\infty\times0$. So, I wrote it as $\frac00$ form and applied L'Hopital. But after first derivative, I again obtained $\frac00$ form. So, I reapplied L'Hopital. But things got quite complicated. So, I abandoned it.

Maybe my approach is wrong. Can you suggest a way? Thanks.

First thing you should do is to change the variable from $x$ to $y=1+\tan x$. — Kavi Rama Murthy, Sep 10 '21 at 06:12
@ZAhmed in the question, fractional part is not mentioned. Also, $(1+\tan x)$ is getting multiplied with the last term (i.e. $1$) as well. — aarbee, Sep 10 '21 at 06:17
@KaviRamaMurthy things indeed got simplified with that substitution. I could easily apply L'Hopital second time now. I'll post my answer below. Thanks. — aarbee, Sep 10 '21 at 06:24

score 3 · Accepted Answer · answered Sep 10 '21 at 06:34

Putting $y=1+\tan x$,

$$L=\lim_{y\to\infty}y(y\ln\frac y{y+1}+1)\\=\lim_{y\to\infty}\frac{y\ln y-y\ln(y+1)+1}{\frac1y}\\=\lim_{y\to\infty}\frac{1+\ln y-\frac{y}{y+1}-\ln(y+1)}{-\frac1{y^2}}\\=\lim_{y\to\infty}\frac{\frac1y-\frac{y+1-y}{(y+1)^2}-\frac1{y+1}}{\frac2{y^3}}\\=\lim_{y\to\infty}\frac{(y+1)^2-y-y(y+1)}{\frac{2(y+1)^2}{y^2}}\\=\lim_{y\to\infty}\frac{y^2+2y+1-y-y^2-y}{2\left(\frac{y+1}y\right)^2}=\frac12$$

So, $4L=2$

score 0 · Answer 2 · answered Sep 10 '21 at 06:53

Use $x=y+\pi/2$, then $$L=\left|\lim_{y\to 0^{-}} \left(\frac{\tan y-1}{\tan y}\right)^2\ln\left(\frac{1-\tan y}{1-2 \tan y}\right)+ \left(\frac{\tan y-1}{\tan y}\right)\right|$$ Let $\tan y=x$, then $$L=\left|\lim_{z\to 0^{-}} \left(\frac{z-1}{z}\right)^2\ln\left(\frac{1-z}{1-2 z}\right)+ \left(\frac{z-1}{z}\right)\right|$$ Use $\log(1-t)=-t-t^2/2+.... |t|<<1$. Then $$L=\left|\lim_{z\to 0^{-}} \left(\frac{z-1}{z}\right)^2[-z-z^2/2+2z+2z^2]+\left(\frac{z-1}{z}\right)\right|$$ $$L=\left|\lim_{z\to 0^-}[(1-z)^2/z+3(1-z)^2/2-1/z+1]\right |=1/2$$

score 0 · Answer 3 · answered Sep 10 '21 at 13:05

We have that by $y=\frac1{1+\tan x}\to 0$

$$L=\left|\lim_{x\to\frac\pi2^-}(1+\tan x)\left[(1+\tan x)\ln\left(\dfrac{1+\tan x}{2+\tan x}\right)+1\right]\right| =\left|\lim_{y\to 0}\frac1y\left[\frac1y\ln\left(\dfrac{1}{1+y}\right)+1\right]\right|$$

and by l'Hospital's rule, Taylor's series or by the method discussed here we have

$$\frac1y\left[\frac1y\ln\left(\dfrac{1}{1+y}\right)+1\right]=\frac{y-\ln\left(1+y\right)}{y^2} \to \frac12$$

therefore $4L=2$.

If $\left|\lim_{x\to\frac\pi2^-}(1+\tan x)\big\{(1+\tan x)\ln\left(\dfrac{1+\tan x}{2+\tan x}\right)+1\big\}\right|$ is $L$, find the value of $4L$.

3 Answers3