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How to determine the limit of the function y=(1-cos^2x)/x^2, if x goes to 0? (don't use...

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axel12 | Student, Undergraduate | eNoter

Posted May 7, 2011 at 2:46 AM via web

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How to determine the limit of the function y=(1-cos^2x)/x^2, if x goes to 0? (don't use l'Hospital)

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giorgiana1976 | College Teacher | Valedictorian

Posted May 7, 2011 at 2:53 AM (Answer #1)

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We'll substitute the difference of the squares from numerator by the product: (1 - cos x)(1 + cos x)

lim (1 - cos x)(1 + cos x)/x^2 = lim [(1 - cos x)/x^2]*lim (1 + cos x)

We'll use the half angle identity:

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

lim 2(sin x/2)^2/x^2*lim (1 + cos x)=2*lim [sin(x/2)/x]*lim [sin(x/2)/x]*lim (1 + cos x)

We'll create the elementary limit:

lim sin x/x = 1

lim [sin(x/2)/x] = lim [sin(x/2)/2*x/2] = (1/2)*lim[sin(x/2)/(x/2)]

lim [sin(x/2)/x] = 1/2

The limit will become:

lim (1 - cos x)(1 + cos x)/x^2 = 2*(1/2)*(1/2)*(1+cos 0)

lim (1 - cos x)(1 + cos x)/x^2 = (1/2)*(1+1)

lim (1 - cos x)(1 + cos x)/x^2 = 1

The requested limit of the function is: lim (1 - cos x)(1 + cos x)/x^2 = 1.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 7, 2011 at 2:59 AM (Answer #2)

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We have to find: lim x-->0 [(1 - (cos x)^2)/x^2]

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

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

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

The Maclaurin series for sin x = x - x^3/3! + x^5/5! - x^7/7! +...

=> lim x--> 0[ ((x - x^3/3! + x^5/5! - x^7/7! +...)/x)^2]

=> lim x--> 0[ ((1 - x^2/3! + x^4/5! - x^6/7! +...))^2]

substitute x = 0

=> (1 - 0 + 0 - 0...)^2

=> 1

The required limit is 1.

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