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.
 A: 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 $$\lim_{x\to 0}\frac{\tan^{n}x-\arctan^{n}x}{x^{n+2}}=\frac{2n}{3}$$
A: 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 $0$.
A: without Taylor series,
you can use 
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ identity.
$$\quad{\lim_{x\to 0}\frac{(\tan x)^{2008}-(\arctan x)^{2008}}{x^{2009}}=\\
\lim_{x\to 0}\frac{(\tan x-\arctan x)(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\
\lim_{x\to 0}\frac{((\tan x-\arctan x))(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\
\lim_{x\to 0}\frac{(\tan x-\arctan x)(\tan^{2007} x+\tan^{2006}x\arctan x+...+\arctan^{2007}x)}{x^{2009}}=\\
\lim_{x\to 0}\frac{(\tan x-\arctan x)(x^{2007} +x^{2006} x^1+...+x^{2007})}{x^{2009}}=\\
\lim_{x\to 0}\frac{(\tan x-\arctan x)2008(x^{2007} )}{x^{2009}}=\\
\lim_{x\to 0}\frac{(\tan x-\arctan x)2008(1 )}{x^2}=\\
\underbrace{\lim_{x\to 0}\frac{(\tan x-\arctan x)2008(1 )}{x^2}=\\}_{\large \text{With respect to "I have a solution using "}\lim_{x\to 0}\frac{\tan x-\arctan x}{x^2}=0} }\\$$
