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Evaluate $$\lim_{x\to 0}\frac{(\tan x)^{2008}-(\arctan x)^{2008}}{x^{2009}}$$ without using Taylor series.

I have a solution using $\lim_{x\to 0}\frac{\tan x-\arctan x}{x^2}=0$, but I would really like to see a different idea.

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  • $\begingroup$ $0{{}}$ surely? $\endgroup$ – Lord Shark the Unknown Sep 22 '17 at 19:30
  • $\begingroup$ Yes, isn't it right? $\endgroup$ – Shroud Sep 22 '17 at 19:37
  • $\begingroup$ Why do you want a different idea when the adapted tool is the Taylor expansion ? $\endgroup$ – Gribouillis Sep 22 '17 at 19:44
  • $\begingroup$ I know the straightforward solution is the one with Taylor series, but I'm just curious to see a beautiful idea. $\endgroup$ – Shroud Sep 22 '17 at 19:47
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    $\begingroup$ I personally prefer the solution which you have indicated in your answer. Any other solution would probably use more sophisticated tools compared to your answer which uses algebra of limits. $\endgroup$ – Paramanand Singh Sep 23 '17 at 0:59
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You can define

$$f(x) = \left(\frac{\tan(x)}{x}\right)^{2008}\qquad g(x) = \left(\frac{\arctan(x)}{x}\right)^{2008}$$

then use $$\frac{f(x)-g(x)}{x} \longrightarrow f^\prime(0) - g^\prime(0)$$

Remark

The following useful inequalities hold

$$|\tan(x) -x| = \int_0^{|x|}\tan^2(t)dt\le \tan^2(x)\int_0^{|x|}dt\le |x| \tan^2(x) \quad \text{for }|x|<\frac{\pi}{2}$$

and $$|\arctan(x) - x| = \int_0^{|x|}\frac{t^2}{1+t^2}dt\le \int_0^{|x|}t^2dt\le \frac{|x|^3}{3}\quad\text{for } x\in{\mathbb R}$$ They imply that $\frac{\tan(x)}{x} - 1 = o(x)$ and $\frac{\arctan(x)}{x}-1 = o(x)$, hence $f^\prime(0)$ and $g^\prime(0)$ exist and are $0$.

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    $\begingroup$ Ingenious! =D Have (+1) and then some. $\endgroup$ – Simply Beautiful Art Sep 22 '17 at 20:03
  • $\begingroup$ You also need to require $f(0)=g(0)=1$ $\endgroup$ – Simply Beautiful Art Sep 22 '17 at 20:05
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    $\begingroup$ +1 for this artistic proof....it looks like a Cezanne's masterpiece $\endgroup$ – Isham Sep 22 '17 at 20:17
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    $\begingroup$ @Isham I'd dream to be a Cezanne of mathematics, but things aren't that easy. Notice that $f^\prime(0)$ and $g^\prime(0)$ are zero without calculation because $f$ and $g$ are even functions. $\endgroup$ – Gribouillis Sep 22 '17 at 20:22
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    $\begingroup$ You can't be no Cezanne if your username is gribouillis $\endgroup$ – Gabriel Romon Sep 23 '17 at 8:33
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The solution you have indicated is the one which is the simplest. But you are perhaps under the impression that it uses Taylor series. This is not the case.


Note that $$\lim_{x\to 0}\frac{\tan x - x} {x^{2}}=\lim_{x\to 0}\frac{\sin x - x\cos x} {x^{2}\cos x}=\lim_{x\to 0}\frac{\sin x - x\cos x} {x^{2}}$$ and then you can split the limit as $$\lim_{x\to 0}\frac{\sin x - x} {x^{2}}+\lim_{x\to 0}x\cdot\frac{1-\cos x} {x^{2}}$$ The second limit is clearly $0\cdot (1/2)=0$ and the the first one is also $0$ via Squeeze Theorem. To apply Squeeze Theorem let's consider the case when $x\to 0^{+}$. Then we have $\sin x <x<\tan x$ which is equivalent to $$\cos x <\frac{\sin x} {x} <1$$ or $$x\cdot\frac{\cos x - 1}{x^{2}}<\frac{\sin x-x} {x^{2}}<0$$ and applying Squeeze gives us the desired result. The case $x\to 0^{-}$ is handled by putting $x=-t$.

Thus we have established that $$\lim_{x\to 0}\frac{\tan x - x} {x^{2}}=0\tag{1}$$ and multiplying this limit with $\lim_{x\to 0}\dfrac{x^{2}}{\tan^{2}x}=1$ we get $$\lim_{x\to 0}\frac{\tan x-x} {\tan^{2}x}=0$$ Putting $x=\arctan t$ and replacing $t$ by $x$ we get $$\lim_{x\to 0}\frac{x-\arctan x} {x^{2}}=0\tag{2}$$ Adding limit equations $(1)$ and $(2)$ we get $$\lim_{x\to 0}\frac{\tan x - \arctan x} {x^{2}}=0\tag{3}$$ The limit in question has got huge exponents as an intimidation tool which we can beat by replacing it with a generic symbol $n$.

We can proceed as follows \begin{align} L &=\lim_{x\to 0}\frac{\tan^{n}x-\arctan^{n}x}{x^{n+1}}\notag\\ &=\lim_{x\to 0}\frac{\tan x - \arctan x} {x^{2}}\cdot\sum_{i=0}^{n-1}\left(\frac{\tan x} {x}\right)^{n-1-i}\left(\frac{\arctan x} {x} \right)^{i} \notag\\ &=0\cdot\sum_{i=0}^{n-1}1\cdot 1=0\notag \end{align}


If the exponent in denominator is $n+2$ then the problem necessitates the use of tools like L'Hospital's Rule and Taylor series to get $$\lim_{x\to 0}\frac{\tan x - \arctan x} {x^{3}}=\frac{2}{3}$$ and as explained above $$\lim_{x\to 0}\frac{\tan^{n}x-\arctan^{n}x}{x^{n+2}}=\frac{2n}{3}$$

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without Taylor series, you can use $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ identity. $$\quad{\lim_{x\to 0}\frac{(\tan x)^{2008}-(\arctan x)^{2008}}{x^{2009}}=\\ \lim_{x\to 0}\frac{(\tan x-\arctan x)(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\ \lim_{x\to 0}\frac{((\tan x-\arctan x))(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\ \lim_{x\to 0}\frac{(\tan x-\arctan x)(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\ \lim_{x\to 0}\frac{(\tan x-\arctan x)(x^{2007} +x^{2006} x^1+...+x^{2007})}{x^{2009}}=\\ \lim_{x\to 0}\frac{(\tan x-\arctan x)2008(x^{2007} )}{x^{2009}}=\\ \lim_{x\to 0}\frac{(\tan x-\arctan x)2008(1 )}{x^2}=\\ \underbrace{\lim_{x\to 0}\frac{(\tan x-\arctan x)2008(1 )}{x^2}=\\}_{\large \text{With respect to "I have a solution using "}\lim_{x\to 0}\frac{\tan x-\arctan x}{x^2}=0} }\\$$

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  • $\begingroup$ You seem to have misread important restraints of the problem at hand. $\endgroup$ – Simply Beautiful Art Sep 22 '17 at 19:40
  • $\begingroup$ How does $\frac{2x^3}{3}$ pop out ? $\endgroup$ – Marios Gretsas Sep 22 '17 at 19:57
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    $\begingroup$ "Agin I used Taylor , but now I don't use it ." could you try again? Are you saying you don't understand what you did? $\endgroup$ – Simply Beautiful Art Sep 22 '17 at 20:03
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    $\begingroup$ The part where you replace all those powers of $\tan x, \arctan x$ by $x^{2007}$ is simply wrong. - 1 $\endgroup$ – Paramanand Singh Sep 23 '17 at 1:03
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    $\begingroup$ @adfriedman: yes I know that. I expect the problem to be fixed with a minor update. But until the update is done it is mathematically wrong. Such steps do not always lead to a correct answer. $\endgroup$ – Paramanand Singh Sep 23 '17 at 4:22

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