Limiting $\tan^{-1}$ at infinity

$\begin{aligned} L & = \lim_{x \to \infty} x \left[ \tan^{-1} \left(\frac {x+1}{x+2} \right) - \tan^{-1} \left(\frac x{x+2} \right) \right] \\ & = \lim_{x \to \infty} x \left[ \tan^{-1} \left(1 - \frac 1{x+2} \right) - \tan^{-1} \left(1 - \frac 2{x+2} \right) \right] & \small \color{#3D99F6} \text{Let }u= \frac 1{x+2} \\ & = \lim_{u \to 0} \left(\frac 1u -2\right) \color{#3D99F6} \left[ \tan^{-1} \left(1 - u \right) - \tan^{-1} \left(1 - 2u \right) \right] & \small \color{#3D99F6} \text{As }\lim_{u \to 0} \left[ \tan^{-1} \left(1 - u \right) - \tan^{-1} \left(1 - 2u \right) \right] = 0 \\ & = \lim_{u \to 0} \frac {\tan^{-1} \left(1 - u \right) - \tan^{-1} \left(1 - 2u \right)}u & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{u \to 0} \frac {- \frac 1{1+(1-u)^2} + \frac 2{1+(1-2u)^2}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }u \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

Using the property that arctan(a)-arctan(b) = arctan(a-b/1+ab)

we get limit as xarctan(x+2/2x^2+5x+4)

As term inside arctan is tending to zero hence arctan(x) tends to x

So limit becomes x^2+2x/2x^2+5x+4

Divide numerator and denominator by x^2 to get limit as 0.5

Expand the arctan functions to the first order in Taylor's expansion.(When we were really solving this problem we should expand to higher terms as well but since we're quite convinced that there "should" be a solution.)

$\displaystyle \arctan { \left( \frac { x+1 }{ x+2 } \right) } =\arctan { \left( \frac { 1+1/x }{ 1+2/x } \right) } \approx \arctan { \left( 1-\frac { 1 }{ x } \right) } \quad when\quad x\quad tends\quad to\quad infinity$ .

The second arctan function is similar.

Therefore,

$\displaystyle \arctan { \left( 1-\frac { 1 }{ x } \right) } =\frac { \pi }{ 4 } +\frac { 1 }{ 1+1 } \left( -\frac { 1 }{ x } \right) =\frac { \pi }{ 4 } +\frac { 1 }{ 2 } \left( -\frac { 1 }{ x } \right)$

$\displaystyle \arctan { \left( \frac { x+1 }{ x+2 } \right) } -\arctan { \left( \frac { x }{ x+2 } \right) } =\arctan { \left( 1-\frac { 1 }{ x } \right) } -\arctan { \left( 1-\frac { 2 }{ x } \right) } =\frac { \pi }{ 4 } +\frac { 1 }{ 2 } \left( -\frac { 1 }{ x } \right) -\left[ \frac { \pi }{ 4 } +\frac { 1 }{ 2 } \left( -\frac { 2 }{ x } \right) \right]$

Hence,

$\displaystyle \lim _{ x\rightarrow \infty }{ x\left[ \arctan { \left( \frac { x+1 }{ x+2 } \right) } -\arctan { \left( \frac { x }{ x+2 } \right) } \right] } =\lim _{ x\rightarrow \infty }{ x\left[ \frac { 1 }{ 2x } \right] } =\frac { 1 }{ 2 } =\boxed{0.5}$

Substitute y=1/x and apply l'hopitals rule.