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

Question

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.

Answer 1

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 $0$.

Answer 2

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.

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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}

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

Answer 3

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