Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
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?
 A: 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
$$\pi -\frac{7}{7500}\approx 3.14066$$
A: 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
$$\lim_{x\to\infty}((x+2)\arctan(x+2)-x\arctan x)=\pi$$
A: 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 $\lim_\limits{x\to 0} \frac{\tan^{-1}x}{x}=1$.
A: 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 $\pi$
A: 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 $\pi$.
