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Calculate the limit (1-cosx)/x^2  , x->0

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creativluana | Student, College Freshman | (Level 1) Honors

Posted December 23, 2010 at 8:10 PM via web

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Calculate the limit (1-cosx)/x^2

 , x->0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 23, 2010 at 8:14 PM (Answer #1)

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We'll substitute the numerator:

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

Lim (1-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2

Lim {2 [sin(x/2)]^2}/x^2 = Lim  [sin(x/2)]^2/(x^2/2)

Lim [sin(x/2)]^2/(x^2/2) = lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)]

Since the elementary limit is:

lim [sin f(x)]/f(x) = 1, if f(x) -> 0, we'll get:

Lim (1-cosx)/x^2 = 1/2

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neela | High School Teacher | (Level 3) Valedictorian

Posted December 23, 2010 at 9:05 PM (Answer #2)

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To find the Lt x-> 0 (1-cosx)/x^2.

Cosx = 1-2 (sinx/2)^2.

1-cosx = 1-(1-(2sinx/2)^2.

1-cosx = 2(sinx/2)^2.

Therefore Lt x--> 0 (1-cosx)/x^2 = Lt x-->0 2(sinx/2)^2/x^2.

 Lt x--> 0 (1-cosx)/x^2 = 2Lt y-->0 (siny)^2/(2y)^2 , where y = x/2.

 Lt x--> 0 (1-cosx)/x^2 = Lt y-->0 (2/4)* [(siny)/y ]^2.

 Lt x--> 0 (1-cosx)/x^2 = 2/4 , as Lt y-->0 (siny)/y = 1.

Therefore  Lt x--> 0 (1-cosx)/x^2 = 1/2.

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lochana2500 | TA , Undergraduate | (Level 1) Valedictorian

Posted January 18, 2013 at 3:12 PM (Answer #3)

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lim x→0 (1-cosx)/x²

= lim x→0 [(1-cosx)(1+cosx)]/[x²(1+cosx)]

= {lim x→0 sinx/x}². {limx→0 1/(1+cosx)}

= 1² . 1/(1+1)

= 1/2

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