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?

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    $\begingroup$ Is L'Hospital's rule applicable? $\endgroup$ – daulomb Jan 2 '14 at 15:26
  • $\begingroup$ 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 $\endgroup$ – gt6989b Jan 2 '14 at 15:27
  • $\begingroup$ wolframalpha.com/input/?i=x%5E3%20*Arctan%5Bx%5D&t=crmtb01 for more info $\endgroup$ – gt6989b Jan 2 '14 at 15:28
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    $\begingroup$ 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. $\endgroup$ – Wintermute Jan 2 '14 at 15:28
  • $\begingroup$ @user40615 yes it is applicable. $\endgroup$ – 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 $1/3$


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