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