# 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?

• See a very similar problem here: math.stackexchange.com/questions/1052932/limit-of-x-lnx – Simon S Dec 5 '14 at 19:28
• A broad hint: let $x=\tan y$; this turns the limit into $\lim_{y\to\pi/2^-}\tan y(\frac\pi2-y)$. Now you can substitute $z=\frac\pi2-y$ and turn the limit into $\lim_{z\to0^+}z\cot z$. Can you see where to go from here? – Steven Stadnicki Dec 5 '14 at 19:33

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
$\lim_{x\to\infty} x(\frac{\pi}{2}-\arctan x)= \lim_{x\to\infty}\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}=\lim_{x\to\infty}\frac{\frac{-1}{1+x^2}}{\frac{-1}{x^2}}=\lim_{x\to\infty}\frac{x^2}{x^2+1}=\cdots$
$$\lim_{x\to\infty}x\left(\frac\pi2-\arctan x\right)=\lim_{x\to\infty}\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}$$
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.