Limit 1

Calculus Level 2

lim ⁡ x → 0 2 − 2 cos ⁡ ( 2 x ) x 2 = ? \lim_{x\rightarrow 0} \dfrac{2-2\cos(2x)}{x^2} = \, ?

1 4 2 5 3

Yosua Sibuea by Yosua Sibuea , 22, Indonesia

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2 solutions

Since cos ⁡ ( 2 x ) = 1 − 2 sin ⁡ 2 ( x ) ⟹ 1 − cos ⁡ ( 2 x ) = 2 sin ⁡ 2 ( x ) \cos(2x) = 1 - 2\sin^{2}(x) \Longrightarrow 1 - \cos(2x) = 2\sin^{2}(x) , the limit can be written as

lim ⁡ x → 0 2 ( 1 − cos ⁡ ( 2 x ) ) x 2 = lim ⁡ x → 0 4 sin ⁡ 2 ( x ) x 2 = 4 ( lim ⁡ x → 0 sin ⁡ ( x ) x ) 2 = 4 \displaystyle\lim_{x \rightarrow 0} \dfrac{2(1 - \cos(2x))}{x^{2}} = \lim_{x \rightarrow 0} \dfrac{4\sin^{2}(x)}{x^{2}} = 4\left(\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x}\right)^{2} = \boxed{4} ,

where we have made use of the identity lim ⁡ x → 0 sin ⁡ ( x ) x = 1 \displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1 .

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Department 8
Feb 4, 2016

The graph above shows that when graph is at 0 the highest point where limit occurs is 4.

Please correct me if you feel I'm wrong.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Obviously , a graphing calculator would always be correct. Why ask for "Please correct me if you feel I'm wrong." ?

Nitesh Chaudhary - 5 years, 4 months ago

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I was wondering whether the statement:

when graph is at 0 the highest point where limit occurs to be 4.

Is correct or not.

Department 8 - 5 years, 4 months ago

Obviously...since f(0 + ^+ )=f(0 − ^- )=f(0)=4..... Hence limit for f(x) exists at 0 and is equal to 4..

Rishabh Jain - 5 years, 4 months ago

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