Vtyjkyuk

$undersetx to inftylim [(x+2)tan^-1(x+2) -x tan^-1x]=?$

My Try :$[(x+2)tan^-1(x+2) -x tan^-1x] = x tan^-1 frac 21+2x+x^2 + 2. tan^-1(x+2)$

$undersetx to inftylim x tan^-1 frac 21+2x+x^2 =0$ [By manipulating L'hospital] and $undersetx to inftylim2. tan^-1(x+2) = pi$

so $undersetx to inftylim [(x+2)tan^-1(x+2) -x tan^-1x]= pi$

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

Yes, the solution is correct.

You are using:
$$arctan apm arctan b=arctanfracapm b1mp ab$$
to simplify:
$$xtan^-1(x+2)-xtan^-1x=xtan^-1frac(x+2)-x1+(x+2)x=xtan^-1frac21+2x+x^2.$$
However, to evaluate the limit you can avoid using the L'Hospital's rule:
$$lim_x to infty x tan^-1 frac 21+2x+x^2=
lim_xtoinfty fractan^-1frac21+2x+x^2frac21+2x+x^2cdot fracfrac21+2x+x^2frac1x=1cdot 0=0,$$
where $lim_limitsxto 0 fractan^-1xx=1$.

Your solution is fine.

Another possible way to do it would be using that, for $t>0$,
$$tan^-1(t)+tan ^-1left(frac1tright)=frac pi 2$$ making
$$A=(x+2)tan^-1(x+2) -x tan^-1(x)$$ $$A=(x+2)left(frac pi 2-tan ^-1left(frac1x+2right) right)-xleft(frac pi 2-tan ^-1left(frac1xright) right)$$
$$A=pi+xtan ^-1left(frac1xright)-(x+2)tan ^-1left(frac1x+2right)$$ and use, for large $t$, the equivalent
$$ttan ^-1left(frac1tright)sim 1$$

If you want to go beyond, you could use Taylor expansions and get
$$A=pi -frac43 x^3+frac4 x^4+Oleft(frac1x^5right)$$

Using your pocket calculator, for $x=10$,
$$A=12 tan ^-1(12)-10 tan ^-1(10)approx 3.14058$$ while the truncated expression gives
$$pi -frac77500approx 3.14066$$

Here is another straightforward way to obtain the limit.

Using the integral relation

$$arctan u=int_0^udtover1+t^2$$

we have

$$beginalign
(x+2)arctan(x+2)-xarctan x
&=2arctan(x+2)+xint_0^x+2dtover1+t^2-xint_0^xdtover1+t^2\
&=2arctan(x+2)+xint_x^x+2dtover1+t^2
endalign$$

Now $2arctan(x+2)topi$ as $xtoinfty$, while

$$0le xint_x^x+2dtover1+t^2le xint_x^x+2dtover x^2=2over xto0$$

Thus

$$lim_xtoinfty((x+2)arctan(x+2)-xarctan x)=pi$$

Using the Mean Value Theorem, the following statement holds:

For $f(z) = 2ztan^-1z$

$$f'(x+xi)=frac2(x+2)tan^-1(x+2)-2xtan^-1xx+2-x$$

For some $xi in (0,2)$

The left hand side simplifies to:

$$f'(x+xi)=2tan^-1(x+xi)+frac2(x+xi)1+(x+xi)^2$$

The right hand side simplifies to the limit you seek.

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

Your solution is fine but it can be shortened. By Lagrange's theorem $(x+2)arctan(x+2)-xarctan(x)$ is twice the value of the derivative of $zarctan z$ at some $zin(x,x+2)$. Since $fracddzzarctan(z) = underbracefracz1+z^2_to 0+underbracearctan(z)_to pi/2$, the wanted limit is $pi$.