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How to evaluate $\lim_{x\to 0+}\dfrac 1x \Big(\dfrac 1{\tan^{-1}x}-\dfrac 1x\Big)$ ? I used L'Hospital's rule but with no success.

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  • $\begingroup$ I wonder, if there is an solution working without L Hospital or the series representation of the series.. In fact there are some nice identities hidden here. $\endgroup$ – Imago Jan 31 '16 at 9:54
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You can use l'Hopital's rule repeatedly. Let $f(x)=x-\tan^{-1}x$ and $g(x)=x^2 \tan^{-1}x.$ Then $$\lim_{x\to 0}\frac {1}{x} (\frac {1}{\tan^{-1}x}-\frac {1}{x})=\lim_{x\to 0}\frac {f(x)}{g(x)}=\lim_{x\to 0}\frac {f'(x)}{g'(x)}.$$ $$\text {Now }\; f'(x)=1-\frac {1}{1+x^2}=\frac {x^2}{1+x^2}$$ $$\text {and } g'(x)=2 x\tan^{-1}x+\frac {x^2}{1+x^2 }=$$ $$=\frac {x}{1+x^2}(\;x+2(1+x^2)\tan^{-1}x )\;).$$ $$\text {So } \quad f'(x)/g'(x)=h(x)/i(x)$$ $$\text {where }\; h(x)=x, \quad i(x)=x+2(1+x^2)\tan^{-1}x.$$ $$\text {We have } \;\lim_{x\to 0}\frac {f'(x)}{g'(x)}= \lim_{x\to 0}\frac {h(x)}{i(x)}=\lim_{x\to 0}\frac {h'(x)}{i'(x)}=\lim_{x\to 0}\frac {1}{(3+4 x \tan^{-1}x)} =1/3.$$

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Maybe it's easier if you unify the denominators

$\lim\limits_{x \to 0^+} \frac{x-arctan(x)}{x^2 arctan(x)} = $

Taylor series of $arctan(x)$ is $ x - \frac{x^3}{3} + \frac{x^5}{5} + ...$

$= \lim\limits_{x \to 0^+} \frac{x-\left[x-\frac{x^3}{3}+o\left(x^3\right)\right]}{x^2 \left[x+o\left(x\right)\right]}$

$= \lim\limits_{x \to 0^+} \frac{\frac{1}{3}x^3+o\left(x^3\right)}{x^3+o\left(x^3\right)}$

$= 1/3$

Probably the $arctan(x)$ in the numerator should be expanded to grade 3, while the one in the denominator should be expanded to grade 1.

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  • $\begingroup$ i did not understand why the negative sign did not apply to $O(x^3)$ $\endgroup$ – Vinay5forPrime Jan 31 '16 at 15:37
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Start with the Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(x^6\right)$$ Perform long division to get $$\frac 1 {\tan^{-1}(x)}=\frac{1}{x}+\frac{x}{3}-\frac{4 x^3}{45}+O\left(x^4\right)$$

I am sure that you can take it from here.

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  • $\begingroup$ Is there any way to get that expansion for $\dfrac 1{\tan^{-1} x}$ without using long-division ? $\endgroup$ – user228168 Jan 31 '16 at 8:44
  • $\begingroup$ This is probably the simplest way (at least, to me). $\endgroup$ – Claude Leibovici Jan 31 '16 at 8:47
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One has

$$\tan^{-1} x=x-{x^3\over 3}+o(x^4)$$

So

$${1\over \tan^{-1} x}-{1\over x}={1\over x}\left({1\over 1-{x^2\over 3}+o(x^3)}-1\right)$$

Now use $1/(1+u)=1-u+o(u)$ to get

$${1\over \tan^{-1} x}-{1\over x}={1\over x}\left(1+{x^2\over 3}+o(x^3)-1\right)={x\over 3}+o(x^2)$$

So the limit we're looking for is $1/3$

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  • $\begingroup$ I think that there is a sign error in the first line. $\endgroup$ – Claude Leibovici Jan 31 '16 at 8:46
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    $\begingroup$ Thanks it's corrected by now. By the way one does not need to go beyond order 3 $\endgroup$ – marwalix Jan 31 '16 at 8:50
  • $\begingroup$ I totally agree that we do not need to go beyond order 3. The problem is that I always want to know how is the limit approached. Have a look at this question of mine (if you are curious) matheducators.stackexchange.com/questions/8339/… . Cheers. $\endgroup$ – Claude Leibovici Jan 31 '16 at 8:53
  • $\begingroup$ You're absolutely right. The asymptotic behaviour is at least as important as the limit $\endgroup$ – marwalix Jan 31 '16 at 9:00
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Let's proceed in the following manner \begin{align} L &= \lim_{x \to 0}\frac{1}{x}\left\{\frac{1}{\tan^{-1} x} - \frac{1}{x}\right\}\notag\\ &= \lim_{x \to 0}\frac{x - \tan^{-1}x}{x^{2}\tan^{-1}x}\notag\\ &= \lim_{t \to 0}\frac{\tan t - t}{t\tan^{2}t}\text{ (putting }x = \tan t)\notag\\ &= \lim_{t \to 0}\frac{\tan t - t}{t^{3}}\cdot\frac{t^{2}}{\tan^{2}t}\notag\\ &= \lim_{t \to 0}\frac{\tan t - t}{t^{3}}\notag\\ &= \lim_{t \to 0}\frac{\sec^{2}t - 1}{3t^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{3}\lim_{t \to 0}\frac{\tan^{2}t}{t^{2}}\notag\\ &= \frac{1}{3}\notag \end{align} We have used the standard limit $\lim\limits_{t \to 0}\dfrac{\tan t}{t} = 1$.

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  • $\begingroup$ That's an awesome solution ! +1 $\endgroup$ – user228168 Feb 1 '16 at 8:36
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    $\begingroup$ @SaunDev: thanks for the compliments! $\endgroup$ – Paramanand Singh Feb 1 '16 at 9:18

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