# 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?

• Is L'Hospital's rule applicable? – daulomb Jan 2 '14 at 15:26
• Are you saying $A, B$ are independent of $x$? Then $A=B=0$ seems like the only possibility, but $x^3 \tan^{-1} (x)$ does not converge I believe – gt6989b Jan 2 '14 at 15:27
• wolframalpha.com/input/?i=x%5E3%20*Arctan%5Bx%5D&t=crmtb01 for more info – gt6989b Jan 2 '14 at 15:28
• If $A=-\frac{\pi}{2}$ and $B$ is anything you want then you will have an indeterminate for which you can begin to apply L'Hopital's rule to. – Wintermute Jan 2 '14 at 15:28
• @user40615 yes it is applicable. – Pumpkin Jan 2 '14 at 15:35

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 $1$.
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.
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 $1/3$