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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?

BigM
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user197895
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  • See a very similar problem here: http://math.stackexchange.com/questions/1052932/limit-of-x-lnx – Simon S Dec 05 '14 at 19:28
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. For some basic information about writing math at this site see e.g. here, here and here. – Tacet Dec 05 '14 at 19:29
  • 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 05 '14 at 19:33

4 Answers4

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

Felix Marin
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$ \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$

BigM
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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.

rae306
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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.

Steven Stadnicki
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