Proving $\lim_{x\to\infty} (x^2 +1)(\frac{\pi}{2} - \arctan{x}) $ doesn't exist. How can I show that $$\lim_{x\to\infty} (x^2 +1)(\frac{\pi}{2} - \arctan{x}) $$ doesn't exist? I used the fact that $$\arctan{x}\ge x-\frac{x^3} {3}, $$ so the initial limit is less than $$\lim_{x\to\infty} \frac{x^5}{3} +O(x^4),$$ therefore the limit tends to infinity. 
Is this enough? If not, then how can I show this rigorously? 
 A: Hint. Observe that, for $x >0$,
$$
\frac{\pi}{2} - \arctan{x}=\arctan{\frac1x}
$$ giving, as $x \to \infty$,
$$
 (x^2 +1)\left(\frac{\pi}{2} - \arctan{x}\right)\ge (x^2+1)\left(\frac1x-\frac1{3x^3} \right).
$$
A: Select $\theta$ such that $tan(\theta)=x$
$$\lim_{x\to\infty} (x^2 +1)(\frac{\pi}{2} - \arctan{x}) $$
$$\lim_{\theta\to \frac{\pi}{2}} tan^2(\theta) +1)(\frac{\pi}{2} - \theta) $$
$$\lim_{\theta\to \frac{\pi}{2}} sec^2(\theta)(\frac{\pi}{2} - \theta) $$
$$\lim_{\theta\to \frac{\pi}{2}} \frac{(\frac{\pi}{2} - \theta)}{cos^2(\theta)} $$
Because both the numerator and denominator head to zero we can employ L'hopitals.
A: If you show that the function is “less than a function having limit $\infty$”, you can't conclude.
Substitute $x=1/t$, using the fact that, for $x>0$,
$$
\frac{\pi}{2}-\arctan x=\arctan\frac{1}{x}
$$
Thus your limit becomes
$$
\lim_{t\to0^+}\frac{(1+t^2)\arctan t}{t^2}=
\lim_{t\to0^+}(1+t^2)\frac{\arctan t}{t}\frac{1}{t}
$$
The first two factors have limit $1$, the last one has limit $\infty$.
