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I need to evaluate the limit of function y=(1-cos2x)/x^2, using trigonometric...

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maisaphie | Student, College Freshman | eNoter

Posted April 15, 2011 at 2:24 AM via web

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I need to evaluate the limit of function y=(1-cos2x)/x^2, using trigonometric identities. x approaches to 0.

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giorgiana1976 | College Teacher | Valedictorian

Posted April 15, 2011 at 2:29 AM (Answer #1)

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We'll use the half angle identity to numerator:

1 - cos 2x = 2 (sin x)^2

We'll substitute 1 - cos 2x by 2 (sin x)^2 and we'll get the equivalent fraction:

(1-cos2x)/x^2 = 2 (sin x)^2/x^2

We'll evaluate the limit of the function 2 (sin x)^2/x^2, if x approaches to 0.

lim 2 (sin x)^2/x^2

We'll create remarkable limit: lim (sin x)/x = 1

According to this, we'll get:

lim 2 (sin x)^2/x^2 = 2lim (sin x)^2/x^2

2lim (sin x)^2/x^2 = 2lim (sin x)/x*lim (sin x)/x

2lim (sin x)^2/x^2 =2*1*1 = 2

The limit of the function y = (1-cos2x)/x^2, if x approaches to 0, is: lim (1-cos2x)/x^2 = 2.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 15, 2011 at 2:32 AM (Answer #2)

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You want the limit of y=(1-cos 2x)/x^2 while x approaches 0.

y = (1-cos 2x)/x^2

=> [1 - (1 - 2*(sin x)^2)]/x^2

=> 2*(sin x)^2/x^2

=> 2*(sin x / x)^2

lim x--> 0 (sin x / x) = 1

Using this identity.

lim x--> 0 [ (1-cos 2x)/x^2]

=> lim x--> 0 (2*(sin x/x)]

=> (2)* lim x--> 0 [(sin x/x)]

=> 2*1

=> 2

The required limit is 2.

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