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Calculus Level 3

lim ⁡ x → 0 1 − cos ⁡ ( 1 − cos ⁡ x ) x 4 = ? \large \lim_{x\to 0} \dfrac{1- \cos(1-\cos x)}{x^4} = \, ?


The answer is 0.125.

Anish Harsha by Anish Harsha , 20, India

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3 solutions

Nihar Mahajan
Feb 11, 2016

lim ⁡ x → 0 1 − cos ⁡ ( 1 − cos ⁡ x ) x 4 = lim ⁡ x → 0 2 ( 1 − cos ⁡ ( 1 − cos ⁡ x ) ) 2 x 4 = lim ⁡ x → 0 2 sin ⁡ 2 ( 1 − cos ⁡ x 2 ) x 4 = lim ⁡ x → 0 2 sin ⁡ 2 ( sin ⁡ 2 ( x / 2 ) ) x 4 ( Note that in above 2 steps , I have excessively used sin ⁡ 2 ( x ) = 1 − c o s ( 2 x ) 2 ) = lim ⁡ x → 0 2 sin ⁡ 2 ( x 2 4 ) x 4 ∵ sin ⁡ ( x ) ≈ x as lim ⁡ x → 0 = lim ⁡ x → 0 2 x 4 16 x 4 ∵ sin ⁡ ( x ) ≈ x as lim ⁡ x → 0 = 2 16 = 1 8 = 0.125 \begin{aligned}\lim_{x\rightarrow 0} \dfrac{1-\cos(1-\cos x)}{x^4} &=\lim_{x\rightarrow 0}\dfrac{2(1-\cos(1-\cos x))}{2x^4} \\ &= \lim_{x\rightarrow 0}\dfrac{2\sin^2\left(\dfrac{1-\cos x}{2}\right)}{x^4} \\ & = \lim_{x\rightarrow 0}\dfrac{2\sin^2(\sin^2(x/2))}{x^4} \\ &\left(\text{Note that in above 2 steps , I have excessively used} \sin^2(x)=\dfrac{1-cos(2x)}{2} \right) \\ &= \lim_{x\rightarrow 0}\dfrac{2\sin^2\left(\dfrac{x^2}{4}\right)}{x^4} \quad\quad\quad \because \ \sin(x) \approx x \text{ as } \lim_{x\rightarrow 0} \\ &= \lim_{x\rightarrow 0} \dfrac{2x^4}{16x^4} \quad\quad\quad \because \ \sin(x) \approx x \text{ as } \lim_{x\rightarrow 0} \\ &= \dfrac{2}{16}=\dfrac{1}{8}=\boxed{0.125}\end{aligned}

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Moderator note:

The 4th equality hasn't been sufficiently justified by " sin ⁡ x ∼ x \sin x \sim x near 0".

We still need to state that " sin ⁡ x \sin x is continuous at x = 0 x = 0 ".

Formally, what you are doing, is claiming that if lim ⁡ g ( x ) = L \lim g(x) = L , then lim ⁡ f ( g ( x ) ) = lim ⁡ f ( L ) \lim f ( g ( x) ) = \lim f (L) . This requires the continuity of f f (and you can easily find a counter example).

The 4th equality hasn't been sufficiently justified by " sin ⁡ x ∼ x \sin x \sim x near 0".

We still need to state that " sin ⁡ x \sin x is continuous at x = 0 x = 0 ".

Formally, what you are doing, is claiming that if lim ⁡ g ( x ) = L \lim g(x) = L , then lim ⁡ f ( g ( x ) ) = lim ⁡ f ( L ) \lim f ( g ( x) ) = \lim f (L) . This requires the continuity of f f (and you can easily find a counter example).

Calvin Lin Staff - 5 years, 3 months ago
Mohamad Zare
Mar 6, 2016

1-Cos(u)~ u u/2 so for 2 time this... . x x x x/x x x x 8=1/8=0.125

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Peter Macgregor
Feb 17, 2016

Using L'Hopital's rule 3 times to successive indeterminate forms reduces the expression to 3 24 = 0.125 \frac{3}{24}=0.125

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you show the working for it? I'm interested to what extent it all cancels out.

My preferred approach in such a situation is to use the Taylor expansion directly instead.

Calvin Lin Staff - 5 years, 3 months ago

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Hi Calvin,

The working is quite long, but as you go along you can informally drop terms by thinking ahead. I'd hate to write it all out in Latex and justify the shortcuts!

Peter Macgregor - 5 years, 3 months ago

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