Trig limit problem: Limit (3x^2)/(2-2cos x) as x approaches 0

[Hoss98]

Hi,
So I've been thinking about this for hours and can't seem to reach the solution

Limit (3x^2)/(2-2cos x) as x approaches 0

I used the identity (1 - cos x)/x as x approaches 0 = 0 and ended up with 3/0 which would be infinity but according to the function's graph, the limit is supposed to be 3 and I can't prove this conclusion.

 

[Deleted member 4993]

Hi,
So I've been thinking about this for hours and can't seem to reach the solution

Limit (3x^2)/(2-2cos x) as x approaches 0

I used the identity (1 - cos x)/x as x approaches 0 = 0 and ended up with 3/0 which would be infinity but according to the function's graph, the limit is supposed to be 3 and I can't prove this conclusion.

Do you know L'Hospital's theorem?

 

[Steven G]

Hi,
So I've been thinking about this for hours and can't seem to reach the solution

Limit (3x^2)/(2-2cos x) as x approaches 0

I used the identity (1 - cos x)/x as x approaches 0 = 0 and ended up with 3/0 which would be infinity but according to the function's graph, the limit is supposed to be 3 and I can't prove this conclusion.

(3x^2)/(2-2cos x)= (3x)/2[(1-cos x)/x]. You lost an x in the numerator. So you get 0/0. As mentioned already, use L'Hospital's theorem.

If you are in Calculus 1, and hence can't use L'Hospital's theorem, then multiply the numerator and denominator by (1+cos x). Then use the fact that sinx/x goes to 1 as x goes to 0