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Is it possible to find $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$$ without l'Hospital's rule?

I have $\lim_{x\to 0}\frac{\tan x}{x}=1$ proved without H. but it doesn't help me with this complicated limit (however I'm sure I have to use it somehow).

I know the answer is $\frac{1}{3}$, so I tried to estimate: $0<\frac{\tan x-x}{x^2\tan x}\le\frac{1}{3}\cdot\frac{\tan x}{x}+g(x)$ with various $g$ and prove that $g(x)\to 0$, but with no result.

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    $\begingroup$ Also without power series? $\endgroup$ – Simon S May 28 '16 at 21:42
  • $\begingroup$ yes, I'm interested in solution without power series also, but estimations taken from power series, which can be proven without the whole theory of power series (like $\sin x\ge x-\frac{x^3}{6}$ for $x\ge 0$) are ok. $\endgroup$ – larry01 May 28 '16 at 21:52
  • $\begingroup$ $\lim_{x\to 0}\frac{\sin x}{x}=1$ and the sine/cosine duplication formulas are enough, but it is quite a tour-de-force. $\endgroup$ – Jack D'Aurizio May 28 '16 at 22:17
  • $\begingroup$ tanx/(x^2 *tanx) -x/(x^2tanx) =1/x^2 -1/(xtanx) . thus we $\endgroup$ – Jacob Wakem May 29 '16 at 17:32
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$$L=\lim_{x\to 0}\frac{x-\tan x}{x^2 \tan x}=\lim_{x\to 0}\frac{\cos x-\frac{\sin x}{x}}{x^2}\cdot\frac{x}{\sin x}=\lim_{x\to 0}\frac{1-2\sin^2\frac{x}{2}-\cos\frac{x}{2}\frac{\sin(x/2)}{(x/2)}}{x^2} $$ gives $L=A+B$ where: $$ A = -\frac{1}{2}+\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2} = -\frac{1}{2}+\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2} = -\frac{1}{2}+\frac{1}{8}=-\frac{3}{8}$$ and $$ B = \lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin(x/2)}{x/2}\right)=\frac{1}{4}\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin x}{x}\right)$$ (assuming such a limit exists) fulfills: $$ 3B = 4B-B = \lim_{x\to 0}\left(\frac{\sin(x/2)}{x/2}-\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{2\sin(x/2)}{\sin (x)}-1\right)$$ so that: $$ B = \frac{1}{3}\lim_{x\to 0}\frac{1}{x^2}\left(\frac{1}{\cos\frac{x}{2}}-1\right)=\frac{1}{3}\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2}=\frac{1}{3}\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2}=\frac{1}{24}$$ and $L=A+B=-\frac{3}{8}+\frac{1}{24}=\color{red}{\large-\frac{1}{3}}$.

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  • $\begingroup$ Great solution! $\endgroup$ – larry01 May 28 '16 at 22:36
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    $\begingroup$ Nice +1. In the step where you write $3B = 4B - B$ you assume that $B$ exists. $\endgroup$ – Paramanand Singh May 29 '16 at 9:33
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If you get to assume that the limit exists, you can show that $$\begin{align}L&=\lim_{x\rightarrow0}\frac{x-\tan x}{x^2\tan x}=\lim_{x\rightarrow0}\frac{-\cos x}{\frac{\sin x}x}\frac{\tan x-x}{x^3}\\ &=-\lim_{x\rightarrow0}\frac{\tan x-x}{x^3}=-\lim_{x\rightarrow0}\frac{\frac{\tan\frac x2-\frac x2}{\left(\frac x2\right)^3}}{4\left(1-\tan^2\frac x2\right)}-\lim_{x\rightarrow0}\frac{\left(\frac{\tan\frac x2}{\frac x2}\right)^2}{4\left(1-\tan^2\frac x2\right)}\\ &=\frac L4-\frac14=-\frac13\end{align}$$ If you don't get to make that assumption, you can derive the first few terms of the series for $\tan^{-1}x$ or $\sin^{-1}x$ algebraically using only $\lim_{x\rightarrow0}\frac{\sin x}x$ and then establish the desired limit, but that is relatively tough sledding.

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first :

$$\lim_{x\to 0}\frac{\tan x-x}{x^3}=\frac{1}{3}$$

proof:Finding $\lim_{x \to 0}\frac{\tan x-x}{x^3}$

now:

$$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}=\lim_{x\to 0}\frac{\tan x-x}{x^3}.\frac{x}{\tan x}=?$$

since :

$$\lim_{x\to 0}\frac{x}{\tan x}=1$$

so : $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}=\frac{1}{3}$$

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