Compute $\displaystyle \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}$
i knew you would come up with something. you are the man! why just multiply the denominator by $\displaystyle \frac 1x$ though? shouldn't we multiply the numerator also?
as always, i'd do the cop out method, and try L'Hopital's rule here (i need to look up the proof for it, it's not fair for me to be using it all the time and not know why it works):
Note that as we try to take the limit, the function goes to $\displaystyle \frac 00$, this fulfils the conditions for using L'Hopital's
Let $\displaystyle L = \lim_{x \to \infty} \frac {\frac {\pi}2 - \arctan x}{\sin \frac 1x}$
Apply L'Hopital's:
$\displaystyle \Rightarrow L = \lim_{x \to \infty} \frac {- \frac 1{x^2 + 1}}{- \frac {\cos \frac 1x}{x^2}}$
$\displaystyle = \lim_{x \to \infty} \frac {\frac {x^2}{x^2 + 1}}{\cos \frac 1x}$
$\displaystyle = \lim_{x \to \infty} \frac {\frac 1{1 + \frac 1{x^2}}}{\cos \frac 1x}$
$\displaystyle = 1$
which confirms Krizalid's answer
$\displaystyle \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}=\lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{x^{-1}\cdot\dfrac{\sin x^{-1}}{x^{-1}}}=\lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{x^{-1}}\cdot\lim_{x\to\infty}\frac{x^{-1}}{\sin x^{-1}}$
We can split the limit because both of them exist (actually they have an indermination, then the entire limit should exists)
The second one is well known (it can be killed with a simple change of variables).
For the first one let's set $\displaystyle u=\arctan x\implies\tan u=x\,\therefore\,u\to\frac\pi2,$ the limit becomes to
$\displaystyle \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{x^{-1}}=\lim_{u\to\pi/2}\frac{\dfrac\pi2-u}{\cot u}$
Since $\displaystyle \cot u=\frac{\cos u}{\sin u}$ and $\displaystyle \cos u=\sin\left(\frac\pi2-u\right),$ the first limit is equal to 1, and $\displaystyle \lim_{x\to\infty}\frac{x^{-1}}{\sin x^{-1}}=1,$ finally
$\displaystyle \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}=1\,\blacksquare$