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Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?

Olivier Oloa
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Kem
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    Get a common denominator and try to use L'Hopital's Rule. Alternately, Maclaurin expand $\arctan(x)$ to a sufficiently high degree to compute the limit. – Ian Aug 28 '15 at 16:18
  • For the latter approach, if you know about the geometric series then you can get the Maclaurin expansion by computing $\int_0^x \frac{1}{1+y^2} dy$ by termwise integration. – Ian Aug 28 '15 at 16:27

5 Answers5

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You may use the Taylor expansion, as $u \to 0$, $$ \frac1{1-u}=1+u+O(u^2) $$ and the Taylor expansion, as $x \to 0$, $$ \arctan x=x-\frac{x^3}3+O(x^5), $$ giving $$ \begin{align} \frac1{x\arctan x}&=\frac1{x(x-\frac{x^3}3+O(x^5))}\\\\ &=\frac1{x^2}\frac1{(1-\frac{x^2}3+O(x^4))}\\\\ &=\frac1{x^2}(1+\frac{x^2}3+O(x^4))\\\\ &=\frac1{x^2}+\frac13+O(x^2) \end{align} $$and

$$ \lim_{x \to 0}\left(\frac1{x\arctan x}- \frac1{x^2}\right)=\frac13. $$

Olivier Oloa
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\begin{align} & \lim_{x\to0} \left( \frac 1 {x \arctan x} - \frac 1 {x^2} \right) = \lim_{x\to0} \frac {x - \arctan x} {x^2 \arctan x} \\[10pt] = {} & \lim_{x\to0} \frac{1 - \frac 1 {1+x^2}}{\frac{x^2}{1+x^2} + 2x \arctan x } \qquad \text{(L'Hopital)} \\[10pt] = {} & \lim_{x\to0} \frac{x^2}{x^2 + 2x(1+x^2) \arctan x} = \lim_{x\to0} \frac{x}{x + 2(1+x^2)\arctan x} \\[10pt] = {} & \lim_{x\to0} \frac 1 {1 + 4x\arctan x + 2} \qquad \text{(L'Hopital)} \\[10pt] = {} & \lim_{x\to 0 } \frac 1 {3+4x\arctan x} = \frac 1 3. \end{align}

Michael Hardy
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Given $$\displaystyle \lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$

Now Let $$\tan^{-1}(x) = y\Rightarrow x=\tan y\;,$$ Then when $x\rightarrow 0,$ Then $y=\tan^{-1}(0)\rightarrow 0$

So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{y\cdot \tan y}-\frac{1}{\tan^2 y} = \lim_{y\rightarrow 0}\frac{\tan y-y}{y\tan^2 y}$$

Now Using $\bf{D-L\; Hopital \; Rule}$

So $$\displaystyle \lim_{y\rightarrow 0}\frac{\sec^2 y-1}{y\cdot 2\tan y\cdot \sec^2 y+\tan^2 y} = \lim_{y\rightarrow 0}\frac{1-\cos^2 y}{2y\tan y+\sin^2 y} = \lim_{y\rightarrow 0}\frac{\sin^2 y}{2y\tan y+\sin^2 y}$$

So limit $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{\frac{2y\tan y}{\sin^2 y}+1} =\lim_{y\rightarrow 0}\frac{1}{\left(\frac{y}{\sin y}\cdot {2}{\cos y}+1\right)}= \lim_{y\rightarrow 0}\frac{1}{2+1} =\frac{1}{3}$$

above we used $$\displaystyle \lim_{y\rightarrow 0}\frac{\sin y}{y} = 1$$

juantheron
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  • I think there is a minor flaw in the last step. Properly speaking, you are proving that $\lim_{y \to 0} \frac{y \tan(y)}{\sin^2(y)}=1$ by splitting it into $\lim_{y \to 0} \frac{y}{\sin(y)}$ and $\lim_{y \to 0} \frac{\tan(y)}{\sin(y)}$. – Ian Aug 28 '15 at 16:40
  • My preceding remark has been corrected. :) – Ian Aug 28 '15 at 16:57
  • Yes Ian i have corrected. – juantheron Aug 28 '15 at 17:13
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Using L'Hôpital and $\lim\limits_{x\to0}\frac{\tan(x)}x=1$, $$ \begin{align} \lim_{x\to0}\left(\frac1{x\tan^{-1}(x)}-\frac1{x^2}\right) &=\lim_{x\to0}\left(\frac{x}{x^3}-\frac{\tan^{-1}(x)}{x^3}\right)\cdot\lim_{x\to0}\frac{x}{\tan^{-1}(x)}\\ &=\lim_{x\to0}\frac{1-\frac1{1+x^2}}{3x^2}\cdot1\\ &=\lim_{x\to0}\frac{\frac{x^2}{1+x^2}}{3x^2}\\ &=\lim_{x\to0}\frac{\frac1{1+x^2}}{3}\\[3pt] &=\frac13 \end{align} $$

robjohn
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Using Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? on,

$$\lim_{x\to0}\dfrac{x-\tan^{-1}x}{x^2\tan^{-1}x}=\lim_{x\to0}\dfrac{\dfrac {x^3}3+O(x^5)}{x^2\tan^{-1}x} =\dfrac13\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}$$ as $x\ne0$ as $x\to0$

Now set $\tan^{-1}x=y\implies x=\tan y,x\to0\implies y\to0$

$$\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}=\lim_{y\to0}\dfrac{\tan y+O(\tan^3y)}{y}=1$$

Community
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lab bhattacharjee
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