Trigonometric Limit without L'Hopital

Question

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

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

Actually not, I am trying to solve it algebraically - thank you for your answer — Antonio Velazquez Bustamante, Jun 24 '15 at 00:57
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 01:20
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 01:35

score 5 · Accepted Answer · answered Jun 24 '15 at 02:41

5

Write the original expression as follows \begin{equation*} \frac{x\cos x-\sin x}{2x^{3}}=\frac{x\cos x-x+x-\sin x}{2x^{3}}=\frac{1}{2}% \left( \frac{\cos x-1}{x^{2}}+\frac{x-\sin x}{x^{3}}\right) . \end{equation*} Now use the standard limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}=-\frac{1}{2},\ \ \ \ \ \ and\ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6} \end{equation*} it follows \begin{equation*} \lim_{x\rightarrow 0}\frac{x\cos x-\sin x}{2x^{3}}=\frac{1}{2}\left( -\frac{1% }{2}+\frac{1}{6}\right) =-\frac{1}{6}. \end{equation*}

answered Jun 24 '15 at 02:41

Idris Addou

4,193

1

In here you can find the proofs w/o l'Hôpital of the first 'standard' limit: http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhospital-rule-or-series-expansion – hjhjhj57 Jun 24 '15 at 02:52
Thank you Idris. It was exactly what I was looking for. – Antonio Velazquez Bustamante Jun 24 '15 at 03:49
You are welcome – Idris Addou Jun 24 '15 at 03:50

score 4 · Answer 2 · answered Jun 24 '15 at 01:52

4

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

answered Jun 24 '15 at 01:52

math110

93,304

This answer you understand? – math110 Jun 24 '15 at 02:31
how do you prove those two inequalities about $\tan x$ without using derivatives? – Paramanand Singh Oct 08 '15 at 10:06

Trigonometric Limit without L'Hopital

2 Answers2