How to compute $\lim_{x\to 0^+}\frac{\arctan x-x}{x^2}$ without Taylor's formula or L'Hôpital's rule?

Question

I have to find $$\lim_{x \rightarrow 0^+} \frac{ \arctan(x)-x}{x^2}$$

without Taylor's formula or L'Hôpital's rule.
How to tackle it? Any idea is welcome.

Answer 1

The function $f(t)$ defined as $\frac{\arctan(t)-t}{t^2}$ for $t\neq 0$ and as $0$ for $t=0$ is an odd function.
To prove that our limit is actually zero, we just need to prove that $f(t)$ is continuous at the origin. We may consider that for any $t>0$ we have $$ t f(t) = -\frac{1}{t}\int_{0}^{t}\frac{x^2}{1+x^2}\,dx $$ hence $$ \left|t f(t)\right| \leq \frac{1}{t}\int_{0}^{t}x^2\,dt = \frac{t^2}{3} $$ and continuity follows from $\left|f(t)\right|\leq\frac{|t|}{3}$ for any $t\in(-\varepsilon,\varepsilon)\setminus\{0\}$.

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Alternative approach: $$ L=\lim_{x\to 0^+}\frac{\arctan x-x}{x^2}=\lim_{\theta\to 0^+}\frac{\theta-\tan\theta}{\tan^2\theta}=\lim_{\theta\to 0^+}\frac{\theta\cos^2\theta-\sin\theta\cos\theta}{\sin^2\theta}$$ or just $$ L=\lim_{x\to 0^+}\frac{\theta\cos\theta-\sin\theta}{\sin^2\theta}=\lim_{x\to 0^+}\frac{\theta\cot\theta-1}{\sin\theta}=\lim_{x\to 0^+}\frac{\theta\cot\theta-1}{\theta}=\lim_{\theta\to 0^+}\left(\cot\theta-\frac{1}{\theta}\right).$$ We may deduce that the limit is zero by proving that $\theta=0$ is a simple pole with residue $1$ of the meromorphic function $\cot\theta$. For such a purpose, we may consider the Weierstrass product of the $\sin$ function, $$ \sin(\theta)=\theta\prod_{n\geq 1}\left(1-\frac{\theta^2}{n^2\pi^2}\right)$$ and by logarithmic differentiation $$ \cot(\theta) = \frac{1}{\theta}+g(\theta) $$ where $g(\theta)$ is a holomorphic function, and an odd function, in a neighbourhood of the origin.

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Yet another way: we may prove that $\lim_{t\to 0}\frac{t-\arctan t}{t^3}=\frac{1}{3}$ through the dominated convergence theorem, since:

$$\lim_{t\to 0}\frac{t-\arctan t}{t^3}=\lim_{t\to 0}\frac{1}{t}\int_{0}^{t}\frac{x^2}{t^2+t^2 x^2}\,dx=\lim_{t\to 0}\int_{0}^{1}\frac{z^2}{1+t^2 z^2}\,dz=\int_{0}^{1}z^2\,dz = \frac{1}{3}.$$

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The last overkill I propose is to exploit squeezing and the following version of the Shafer-Fink inequality

$$\forall x\in(0,1),\qquad \arctan(x)\approx\frac{3x}{1+2\sqrt{1+x^2}}$$ that can be proved through the strong convexity of the cotangent function in a right neighbourhood of the origin and its duplication formulas (just have a look at the linked article).
And you are free to replace such approximation by a better one provided by the Gauss continued fraction, of course.

Answer 2

Note that

$$\begin{align} \arctan x-x &= \int_0^x\left({1\over1+t^2}-1\right)dt\\ &=\int_0^x{-t^2\over1+t^2}dt\\ &=\int_0^1{-(xu)^2\over1+(xu)^2}xdu\\ &=-x^3\int_0^1{u^2\over1+x^2u^2}du \end{align}$$

Hence

$$0\le\left|\arctan x-x\over x^2\right|=\left|x\int_0^1{u^2\over1+x^2u^2}du\right|\le\left|x\int_0^1u^2du\right|={1\over3}|x|\to0\text{ as }x\to0$$

So by the squeeze theorem

$$\lim_{x\to0}\left|\arctan x-x\over x^2\right|=0$$

(Remark: This is basically the same idea as in Jack D'Aurizio's answer, just exposited somewhat differently.)

Answer 3

This one is not that hard like $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6}$$ By a substitution $\arctan x = t$ and the fact that $(\tan t)/t \to 1$ as $t \to 0$ it is easily seen that the problem is equivalent to proving that $$\lim_{t \to 0}\frac{\tan t - t}{t^{2}} = 0\tag{1}$$ and it is this relation which we will establish.

Let us first deal with $t \to 0^{+}$. Then we know that if $0 < t < \pi/2$ then $$\sin t < t < \tan t$$ and hence we have $$0 < \frac{\tan t - t}{t^{2}} < \frac{\tan t - \sin t}{t^{2}} = \frac{(1 - \cos t)\sin t}{t^{2}\cos t}$$ and applying Sqeeze theorem we can see that $$\lim_{t \to 0^{+}}\frac{\tan t - t}{t^{2}} = 0$$ For $t \to 0^{-}$ we can just put $t = -y$ and let $y \to 0^{+}$. We have thus established the limit formula $(1)$.

Answer 4

The following should prove that the limit is $0$ if it exists.

Dot he change of variables $x = 2t/(1-t^2)$, where $t \to 0^+$. Because $2\arctan(t) = \arctan(2t/(1-t^2)),$ we get the expression $$\frac{2\arctan(t)-2t/(1-t^2)}{(2t/(1-t^2))^2} = \frac{1}{2}\frac{\arctan(t)-t}{t^2}-\arctan(t)+\frac{1}{2}\arctan(t)t^2+\frac{1}{2}t.$$

Here the first term is $1/2$ times the original expression and the other terms tend to $0$.

Thus we have $$\lim_{x\to 0^+}\frac{\arctan(x)-x}{x^2} = \frac{1}{2}\lim_{x\to0^+}\frac{\arctan(x)-x}{x^2}.$$

It's not a complete answer but didn't fit in a comment.

Repeating the same argument we can show for $n \geq 1$, that

$$\lim_{x\to 0^+}\frac{\arctan(x)-x}{x^2} = \frac{1}{2^n}\lim_{x\to0^+}\frac{\arctan(x)-x}{x^2}.$$

On second thoughts, maybe it works. Without being too formal, the above argument shows that, by induction, we can estimate $|(\arctan(x)-x)/x^2|$ by $(1/2)^n|(\arctan(x)-x)/x^2|$ around the origin. Letting $n \to \infty$ ought to show that the limit actually exists and tends to zero. I welcome any comments on this.

Answer 5

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\forall\ x > 0\ \exists\ \xi\ \mid\ 0 < \xi < x\quad\mbox{and}\quad {x - \arctan\pars{x} \over x^{2}} = {1 \over x}\,\pars{1 - {1 \over \xi^{\, 2} + 1}} = {1 \over x}\,{\xi^{\, 2} \over \xi^{\, 2} + 1} \\[5mm] &\ \mbox{Also}\,,\ 0 < {1 \over x}\,{\xi^{\, 2} \over \xi^{\, 2} + 1} < {\xi \over \xi^{\, 2} + 1}\quad\implies\quad \color{#f00}{\lim_{x \to 0^{+}}{\arctan\pars{x} - x \over x^{2}}} = \color{#f00}{0}\quad \pars{~Sandwich\ Theorem~} \end{align}