lim ⁡ x → 0 ( 2 x − sin ⁡ 2 x ) / x 3 \lim_{x\to 0}(2x-\sin 2x)/x^3

Calculus Level 2

lim ⁡ x → 0 2 x − sin ⁡ 2 x x 3 = ? \large \lim_{x \to 0} \frac {2x-\sin 2x}{x^3} = ?


The answer is 1.3333333.

Aly Ahmed by Aly Ahmed , 34, Egypt

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2 solutions

Chew-Seong Cheong
Sep 26, 2018

Relevant wiki: Maclaurin Series

L = lim ⁡ x → 0 2 x − sin ⁡ 2 x x 3 By Maclaurin series = lim ⁡ x → 0 2 x − ( 2 x − ( 2 x ) 3 3 ! + ( 2 x ) 5 5 ! − ⋯ ) x 3 = lim ⁡ x → 0 8 x 3 3 ! − 32 x 5 5 ! + 128 x 7 7 ! − ⋯ x 3 Divide up and down by x 3 = lim ⁡ x → 0 8 6 − 32 x 2 5 ! + 128 x 5 7 ! − ⋯ 1 = 4 3 ≈ 1.333 \begin{aligned} L & = \lim_{x\to 0} \frac {2x-\color{#3D99F6}\sin 2x}{x^3} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x\to 0} \frac {2x - \color{#3D99F6}\left(2x - \frac {(2x)^3}{3!} + \frac {(2x)^5}{5!} - \cdots \right)}{x^3} \\ & = \lim_{x\to 0} \frac {\frac {8x^3}{3!} - \frac {32x^5}{5!} + \frac {128x^7}{7!} - \cdots}{x^3} & \small \color{#3D99F6} \text{Divide up and down by }x^3 \\ & = \lim_{x\to 0} \frac {\frac 86 - \frac {32x^2}{5!} + \frac {128x^5}{7!} - \cdots}1 \\ & = \frac 43 \approx \boxed{1.333} \end{aligned}

7 Helpful 0 Interesting 2 Brilliant 0 Confused
Blan Morrison
Sep 26, 2018

Relevant wiki: L'Hopital's Rule - Basic

L = 2 x − sin ⁡ ( 2 x ) x 3 L=\frac{2x-\sin(2x)}{x^3} lim ⁡ x → 0 L = 2 x − sin ⁡ ( 2 x ) x 3 \displaystyle\lim_{x\rightarrow 0}L=\frac{2x-\sin(2x)}{x^3} = 2 − 2 cos ⁡ ( 2 x ) 3 x 2 =\frac{2-2\cos(2x)}{3x^2} = 4 sin ⁡ ( 2 x ) 6 x =\frac{4\sin(2x)}{6x} = 8 cos ⁡ ( 2 x ) 6 = 8 6 = 1.333 … =\frac{8\cos(2x)}{6}=\frac{8}{6}=\boxed{1.333\dots}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I don't know why this happens, but if there is any value besides 2 x 2x , the limit goes to infinity, but L'Hopital's says otherwise. Can someone please explain?

Blan Morrison - 2 years, 8 months ago

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I don't understand what you mean. I googled "graph (3x-sin(3x))/x^3" and at '0' it is 4.5 (9/2), which is precisely what L'Hopital's predicts.

I tried "graph (4x-sin(4x))/x^3" and "graph (5x-sin(5x))/x^3" and they too returned a finite value at '0'.

PS: Don't just change one of the "2x". If for instance you do ( 3 x − s i n ( 2 x ) ) x 3 \frac{\left(3x-sin(2x)\right)}{x^3} , on the 1st derivative you get ( 3 − 2 c o s ( 2 x ) ) 3 x 2 \frac{\left(3-2cos(2x)\right)}{3x^2} which evaluated at '0' is 1 0 \frac{1}{0} , and therefore you can not continue to apply L'Hopital.

P S - 2 years, 6 months ago

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That's where I went wrong; thank you! Two months ago, I didn't realize it applied only to 0/0 and ∞/∞.

Blan Morrison - 2 years, 6 months ago

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