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

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.

• $0{{}}$ surely? Commented Sep 22, 2017 at 19:30
• Yes, isn't it right? Commented Sep 22, 2017 at 19:37
• Why do you want a different idea when the adapted tool is the Taylor expansion ? Commented Sep 22, 2017 at 19:44
• I know the straightforward solution is the one with Taylor series, but I'm just curious to see a beautiful idea. Commented Sep 22, 2017 at 19:47
• 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. Commented Sep 23, 2017 at 0:59

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 $0$.

• Ingenious! =D Have (+1) and then some. Commented Sep 22, 2017 at 20:03
• You also need to require $f(0)=g(0)=1$ Commented Sep 22, 2017 at 20:05
• +1 for this artistic proof....it looks like a Cezanne's masterpiece Commented Sep 22, 2017 at 20:17
• @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. Commented Sep 22, 2017 at 20:22
• You can't be no Cezanne if your username is gribouillis Commented Sep 23, 2017 at 8:33

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 $$\lim_{x\to 0}\frac{\tan^{n}x-\arctan^{n}x}{x^{n+2}}=\frac{2n}{3}$$

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

• You seem to have misread important restraints of the problem at hand. Commented Sep 22, 2017 at 19:40
• How does $\frac{2x^3}{3}$ pop out ? Commented Sep 22, 2017 at 19:57
• "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? Commented Sep 22, 2017 at 20:03
• The part where you replace all those powers of $\tan x, \arctan x$ by $x^{2007}$ is simply wrong. - 1 Commented Sep 23, 2017 at 1:03
• @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. Commented Sep 23, 2017 at 4:22