Answer
$0$
Work Step by Step
Introduce a new variable $t=e^{x}$.
When $ x\rightarrow\infty$, we also have $ t\rightarrow\infty$
$\displaystyle \lim_{x\rightarrow\infty}\frac{e^{x}\tan^{-1}e^{x}}{e^{2x}+x}=\lim_{t\rightarrow\infty}\frac{t\tan^{-1}t}{t^{2}+\ln t}$
Rewrite as a product of limits,
$=\displaystyle \lim_{t\rightarrow\infty}\frac{\tan^{-1}t}{t}\cdot\lim_{t\rightarrow\infty}\frac{t^{2}}{t^{2}+\ln t}=L_{1}\cdot L_{2}$
$L_{1}=\displaystyle \lim_{t\rightarrow\infty}\frac{\tan^{-1}t}{t}$=$\qquad \left[\begin{array}{ll}
\text{numerator:} & \tan^{-1}t\rightarrow\frac{\pi}{2}\\
\text{denominator:} & t\rightarrow\infty
\end{array}\right]\qquad=0$
The product $L_{1}\cdot L_{2}$ is zero, if $L_{2}$ is defined
We check if $L_{2}$ is defined:
$L_{2}=\displaystyle \lim_{t\rightarrow\infty}\frac{t^{2}}{t^{2}+\ln t}\qquad $... $\displaystyle \frac{\infty}{\infty}$, apply L'Hopital's rule
$L_{2}=\displaystyle \lim_{t\rightarrow\infty}\frac{2t}{2t+\frac{1}{t}}\qquad $divide with $\displaystyle \frac{t}{t}$
$L_{2}=\displaystyle \lim_{t\rightarrow\infty}\frac{2}{2+\frac{1}{t^{2}}}\qquad [\frac{1}{t^{2}}\rightarrow 0]$
$L_{2}=1$
$L=L_{1}\cdot L_{2}=0\cdot 1=0$
