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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?

Martin Sleziak
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    L'Hospital $\frac{\pi/2 - \arctan(x)}{1/(x^2+1)}$ leads you to compute the limit of $(-1/(1+x^2))/(-2x/(x^2+1)^2)=(x^2+1)^2/(2x(x^2+1))$, which $\to\infty$ because the degree of the numerator is $4$ and that of the denominator is $3$. –  May 01 '18 at 21:12
  • It seems, however, that $\lim_{x\to\infty} \frac{(x^2 +1)(\frac{\pi}{2} - \arctan{x})}{x}=1$ – Steven Alexis Gregory May 01 '18 at 21:20
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    Thanks everyone! I understood. Sadly I cant accept more than 1 answer. –  May 01 '18 at 21:50

3 Answers3

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Hint. Observe that, for $x >0$, $$ \frac{\pi}{2} - \arctan{x}=\arctan{\frac1x} $$ giving, as $x \to \infty$, $$ (x^2 +1)\left(\frac{\pi}{2} - \arctan{x}\right)\ge (x^2+1)\left(\frac1x-\frac1{3x^3} \right). $$

Olivier Oloa
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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.

Mason
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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 $\infty$.

egreg
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