Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$

Question

$\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$

My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$

$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$

so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$

Can anyone please correct me If I have gone wrong anywhere?

Answer 1

Yes, the solution is correct.

You are using: $$\arctan a\pm \arctan b=\arctan\frac{a\pm b}{1\mp ab}$$ to simplify: $$x\tan^{-1}(x+2)-x\tan^{-1}x=x\tan^{-1}\frac{(x+2)-x}{1+(x+2)x}=x\tan^{-1}\frac{2}{1+2x+x^2}.$$ However, to evaluate the limit you can avoid using the L'Hospital's rule: $$\lim_{x \to \infty} x \tan^{-1} \frac {2}{1+2x+x^2}= \lim_{x\to\infty} \frac{\tan^{-1}\frac{2}{1+2x+x^2}}{\frac{2}{1+2x+x^2}}\cdot \frac{\frac{2}{1+2x+x^2}}{\frac{1}{x}}=1\cdot 0=0,$$ where $\lim_\limits{x\to 0} \frac{\tan^{-1}x}{x}=1$.

Answer 2

Using the Mean Value Theorem, the following statement holds:

For $f(z) = 2z\tan^{-1}{z}$

$$f'(x+\xi)=\frac{2(x+2)\tan^{-1}{(x+2)}-2x\tan^{-1}{x}}{x+2-x}$$

For some $\xi \in (0,2)$

The left hand side simplifies to:

$$f'(x+\xi)=2\tan^{-1}{(x+\xi)}+\frac{2(x+\xi)}{1+(x+\xi)^2}$$

The right hand side simplifies to the limit you seek.

By taking the limit of the left hand side, the limit is clearly $\pi$

Answer 3

Your solution is fine.

Another possible way to do it would be using that, for $t>0$, $$\tan^{-1}(t)+\tan ^{-1}\left(\frac{1}{t}\right)=\frac \pi 2$$ making $$A=(x+2)\tan^{-1}(x+2) -x \tan^{-1}(x)$$ $$A=(x+2)\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x+2}\right) \right)-x\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x}\right) \right)$$ $$A=\pi+x\tan ^{-1}\left(\frac{1}{x}\right)-(x+2)\tan ^{-1}\left(\frac{1}{x+2}\right)$$ and use, for large $t$, the equivalent $$t\tan ^{-1}\left(\frac{1}{t}\right)\sim 1$$

If you want to go beyond, you could use Taylor expansions and get $$A=\pi -\frac{4}{3 x^3}+\frac{4}{ x^4}+O\left(\frac{1}{x^5}\right)$$

Using your pocket calculator, for $x=10$, $$A=12 \tan ^{-1}(12)-10 \tan ^{-1}(10)\approx 3.14058$$ while the truncated expression gives $$\pi -\frac{7}{7500}\approx 3.14066$$

Answer 4

Here is another straightforward way to obtain the limit.

Using the integral relation

$$\arctan u=\int_0^u{dt\over1+t^2}$$

we have

$$\begin{align} (x+2)\arctan(x+2)-x\arctan x &=2\arctan(x+2)+x\int_0^{x+2}{dt\over1+t^2}-x\int_0^x{dt\over1+t^2}\\ &=2\arctan(x+2)+x\int_x^{x+2}{dt\over1+t^2} \end{align}$$

Now $2\arctan(x+2)\to\pi$ as $x\to\infty$, while

$$0\le x\int_x^{x+2}{dt\over1+t^2}\le x\int_x^{x+2}{dt\over x^2}={2\over x}\to0$$

Thus

$$\lim_{x\to\infty}((x+2)\arctan(x+2)-x\arctan x)=\pi$$

Answer 5

Your solution is fine but it can be shortened. By Lagrange's theorem $(x+2)\arctan(x+2)-x\arctan(x)$ is twice the value of the derivative of $z\arctan z$ at some $z\in(x,x+2)$. Since $\frac{d}{dz}z\arctan(z) = \underbrace{\frac{z}{1+z^2}}_{\to 0}+\underbrace{\arctan(z)}_{\to \pi/2}$, the wanted limit is $\pi$.