How to evaluate the following limit? $\lim\limits_{x\to\infty}x\left(\frac\pi2-\arctan x\right).$ How can I evaluate this limit:
$$\lim_{x\to\infty}x\left(\frac\pi2-\arctan x\right).$$
I know that the correct answer is $1$, but why?
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\begin{align}&\color{#66f}{\large%
\lim_{x\ \to\ \infty}x\ \overbrace{\bracks{{\pi \over 2} - \arctan\pars{x}}}
^{\ds{\color{#c00000}{{\pi \over 2} - {\pi \over 2}\,\sgn\pars{x}+
                      \arctan\pars{1 \over x}}}}}\ =\
\lim_{x\ \to\ \infty}\bracks{x\arctan\pars{1 \over x}}
\\[5mm]&=\lim_{x\ \to\ 0}{\arctan\pars{x} \over x}
=\lim_{x\ \to\ 0}{1/\pars{x^{2} + 1} \over 1}=\color{#66f}{\Large 1}
\end{align}
A: $ \lim_{x\to\infty} x(\frac{\pi}{2}-\arctan x)= \lim_{x\to\infty}\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}=\lim_{x\to\infty}\frac{\frac{-1}{1+x^2}}{\frac{-1}{x^2}}=\lim_{x\to\infty}\frac{x^2}{x^2+1}=\cdots$
A: Hint:
$$\lim_{x\to\infty}x\left(\frac\pi2-\arctan x\right)=\lim_{x\to\infty}\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}$$
Now apply L'Hôpitals Rule.
A: To flesh my comment out into a (hinting) answer: this is straightforward to do without L'Hôpital, as long as you know some classic trig limits.  First, make the substitution $y=\arctan x$ (note that for the limit to have the value you're talking about, we need to be using the principal branch of the arctangent both here and in the original question — i.e., the one whose range runs from $-\frac\pi2$ to $\frac\pi2$).  Since $\tan y$ goes to (positive) infinity as $y$ approaches $\frac\pi2$ from below, this turns the limit into $\lim\limits_{y\to\frac\pi2^-} \tan y(\frac\pi2-y)$.  Now, we can perform another substitution, $z=\frac\pi2-y$; since $y$ was approaching $\frac\pi2$ from below, $z$ will approach $0$ from above.  Using $\cot x=\tan(\frac\pi2-x)$, this transforms the limit into $\lim\limits_{z\to0^+}z\cot z$.  Now, you can just use the definition of cotangent — along with a trig limit you should be very familiar with — to compute the final value.
