Calculus: Limit (cos(x)-1)/x as x tends to 0

Professor Zap proves that (cos(x)-1)/x tends to 0 as x tends to 0.

Category:

Education

Tags:

• Calculus
• Limits
• Trig
• Trigonometry
• Math
• Mathematics
• University of South Alabama

[kimberlyssy]

@Fullmetal1222 this is late.... but cos^2(x)-1 is not equal to sin^2(x). 1-cos^2(x) is equal to sin^2(x). So you factor -1 from cos^2(x)-1 to get -1(-cos^2(x) +1) which is just -1(sin^2(x)).

[Fullmetal1222]

why is there a -1 sin^2x on top when cos^2-1 equals sin^2x??

[zyzzyzus]

Using this and the one for sin(x)/x you can prove the derivative of sin(x) is cos(x): The definition of a derivative is the limit as h->0 of (f(x+h)-f(x))/h So for sin(x) it will be (sin(x+h)-sin(x))/h = [sin(x)cos(h) + cos(x)sin(h) - sin(x)]/h = [sin(x) (cos(h)-1) + cos(x)sin(h)]/h This can be split into two parts: sin(x) * [cos(h) - 1]/h and cos(x) * sin(h)/h So taking the limit as h->0 for [cos(h) - 1]/h gives 0 and for sin(h)/h it is 1. sin(x)*0 + cos(x)*1 = cos(x) These are good videos.

[Rikiicherry]

Amazing Thank you :D

[ivanrengo83]

thanks!! your explanation was very helpful. greetings from Argentina

[ScreamitsShirley]

You are helpful thank you

[kaibzolis]

this is (ah) Eff of Ecks plus Delta Ecks (ah) Minus eff of ecks over delta ecks!