Do you remember this limit?

Calculus Level 4

lim ⁡ x → 0 1 − cos ⁡ 2 x 2 x = ? \large \lim_{x\to 0} \dfrac{\sqrt{1-\cos2x}}{\sqrt2 x} = \, ?

0 Does not exist 2 1

Suneel Chauhan by Suneel Chauhan , 22, Pakistan

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1 solution

Suneel Chauhan
May 10, 2016

1 − cos ⁡ 2 x = 2 sin ⁡ 2 x 1 - \cos 2x = 2 \sin^2 x
Thefore it becomes 2 ∣ sin ⁡ x ∣ 2 x = ∣ sin ⁡ x ∣ x \dfrac{\sqrt{2} \lvert \sin x \rvert }{\sqrt{2} x} = \dfrac{\lvert \sin x \rvert}{ x}

Taking LHL, lim ⁡ x → 0 + ∣ sin ⁡ x ∣ x = 1 \displaystyle \lim _{x \to 0^+} \dfrac{\lvert \sin x \rvert }{x} =1
Taking RHL, lim ⁡ x → 0 − ∣ sin ⁡ x ∣ x = − 1 \displaystyle \lim _{x \to 0^-} \dfrac{\lvert \sin x \rvert }{x} =-1
RHL does not equal to LHL, therefore limit does exist not.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for sharing your insight. I have edited the Latex so the math equations are easier to read.

Pranshu Gaba - 5 years, 1 month ago

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