# Trigonometric Limit without L'Hopital

I am having problems solving this limit without L'Hopital or series. $$\lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2 x^3}$$

I tried some trigonometric manipulations without success. I tried Trigonometric identities with no luck and separating $$\frac{x\cos(x)}{2 x^3} and \frac{sin(x)}{2 x^3}$$ lead me nowhere, each of this limits are infinity. I kow the result is

$$\lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2 x^3} = \frac{-1}{6}$$

• You wanna use Taylor series? – Braindead Jun 24 '15 at 0:55
• Actually not, I am trying to solve it algebraically - thank you for your answer – Antonio Velazquez Bustamante Jun 24 '15 at 0:57
• Well, you will have to use some properties of $\sin$ and $\cos$ - which would you accept as algebraic, if you are reluctant to use the first few term of the Taylor series? – Damian Reding Jun 24 '15 at 1:20
• Thank you. I tried some trigonometric manipulation without success. The idea is to avoid the indetermination without l hopital or series. – Antonio Velazquez Bustamante Jun 24 '15 at 1:35

since $$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=\lim_{x\to 0}\dfrac{-\cos{x}(\tan{x}-x)}{2x^3}=-\lim_{x\to 0}\dfrac{\tan{x}-x}{2x^3}$$ Use this two inequality $$\tan{x}>x+\dfrac{1}{3}x^3\Longrightarrow \dfrac{\tan{x}-x}{2x^3}>\dfrac{1}{6},x\in [0,\dfrac{\pi}{2})$$ Other hand $$\tan{x}<\dfrac{3x}{3-x^2}\Longrightarrow \dfrac{\tan{x}-x}{2x^3}<\dfrac{1}{2(3-x^2)},x\in [0,\dfrac{\pi}{2})$$ so $$\lim_{x\to\infty}\dfrac{\tan{x}-x}{2x^3}=\dfrac{1}{6}$$ so $$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=-\dfrac{1}{6}$$
• how do you prove those two inequalities about $\tan x$ without using derivatives? – Paramanand Singh Oct 8 '15 at 10:06