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.
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

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

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