Finding limit by l'hospital rule. 
Q. Find the constants $A$ and $B$ such that $\displaystyle\lim\limits_{x\to\infty}x^3\left(A+\dfrac Bx+\arctan x\right)$ exists. Calculate the limit.

How can I find the limit by l'hospital rule?
 A: We have
$$\frac{A+(B/x)+\arctan x}{1/{x^3}}.$$
Since 
$$\lim_{x\to\infty}\arctan x=\frac{\pi}{2}, \lim_{x\to\infty}\frac{B}{x}=0,$$
$A$ has to be $-\pi/2.$
In addition to this, we have
$$\frac{-Bx^{-2}+\{1/(1+x^2)\}}{-3x^{-4}}=\cdots=\frac{(B-1)x^2+B}{3+(3/{x^2})}.$$
This implies that $B$ has to be $1$.
A: HINT: To use L'Hopital's Rule on this question, change the given expression to $$\lim _{x\to\infty} \frac{x^3}{\frac{1}{A + \frac{B}{X} + \arctan(x)}}$$
or $$\lim_{x\to\infty} \frac{A + \frac{B}{X} + \arctan(x)}{1/x^3}$$ As mathlove has done.
A: If you want that the limit exists and is finite, then you can observe that, for $x>0$, we have
$$
\arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}
$$
so your limit, after the transformation $t=1/x$, becomes
$$
\lim_{t\to 0^+}\frac{A+Bt+\pi/2-\arctan t}{t^3}.
$$
In order that the limit be finite, you need
$$
A+B0+\frac{\pi}{2}-\arctan0=0
$$
that is $A=-\frac{\pi}{2}$. Thus the limit becomes
$$
\lim_{t\to 0^+}\frac{Bt-\arctan t}{t^3}\overset{(H)}{=}
\lim_{t\to 0^+}\frac{B-1/(1+t^2)}{3t^2}=
\lim_{t\to 0^+}\frac{Bt^2+B-1}{3t^2(1+t^2)}
$$
Thus …

 $B=1$ and the limit is $1/3$

