limit with $\arctan$ and $\ln$ I have to find the limit 
and want ask about a hint:
$$\lim_{x \to \infty} \frac{\frac{1}{2}\pi-\text{tan}^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)}$$
I dont have idea what to do. Maple show me that the answer is $\infty$.
 A: Notice that
$$\arctan x+\arctan\left(\frac1x\right)=\frac\pi2,\quad x>0$$
so let $y=\frac 1x$ then the limit becomes
$$\lim_{y\to0}\frac{\arctan y}{\ln(1+y^2)}=\lim_{y\to0}\frac{y}{y^2}=\infty$$
A: HINT:
From  $\#11$ of this, $\displaystyle \tan^{-1}x+\cot^{-1}x=\frac\pi2$ for $x\ge0$
$$F=\lim_{x \to \infty} \frac{\frac{1}{2}\pi-\tan^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)} =\lim_{x \to \infty} \frac{\cot^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)}$$
Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
$$F=\lim_{x \to \infty} \frac{\tan^{-1}\frac1x}{\ln\left(1+\frac{1}{x^2}\right)}$$
Setting $x=\frac1h$
$$F=\lim_{h\to0}\frac{\tan^{-1}h}{\ln(1+h^2)}=\lim_{h\to0}\frac{\tan^{-1}h}h\cdot\frac1{\lim_{h\to0}\dfrac{\ln(1+h^2)}{h^2}}\cdot\lim_{h\to0}\frac1h$$
A: What about a little l'Hospital's Rule?
$$\lim_{x\to\infty}\frac{\frac\pi2-\arctan x}{\log\left(1+\frac1{x^2}\right)}\stackrel{\text{l'H}}=\lim_{x\to\infty}-\;\left(-\frac x2\right)=\infty$$
A: You can also just use the l'Hôpital's rule:
$$\lim_{x\to\infty}\frac{\frac{1}{2}\pi-\text{tan}^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)}=\lim_{x\to\infty}\frac{-\frac1{1+x^2}}{\frac{-\frac2{x^3}}{1+\frac1{x^2}}}=\lim_{x\to\infty}\frac1{1+x^2}\cdot\frac{x^3+x}2=\infty$$
