# limits problem

• May 3rd 2009, 07:33 AM
manalive04
limits problem
compute limit

lim cos(3x)-1
x->0 sin(x)sin(x)
• May 3rd 2009, 07:42 AM
Rapha
Hi!

Quote:

Originally Posted by manalive04
compute limit

lim cos(3x)-1
x->0 sin(x)sin(x)

$\lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)} = - \frac{9}{2}$

Yours, Rapha
• May 3rd 2009, 07:58 AM
manalive04
Quote:

Originally Posted by Rapha
Hi!

$\lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)} = - \frac{9}{2}$

Yours, Rapha

how did u do that????
• May 3rd 2009, 08:13 AM
Rapha
Okay, it is

$\lim_{x \to 0} \ cos(3x)-1 = 1-1 = 0$

$\lim_{x \to 0} \ sin(x)sin(x) = 0$

Therefor we use L'Hospital

We know that $cos(3x) = 4cos(x)^3 - 3cos(x)$

Thus

$cos(3x)-1 = 4cos(x)^3 - 3cos(x) =:f(x)$

=> $f'(x) = - 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)$

$g(x):= sin(x)sin(x)$

$g'(x) = 2sin(x)cos(x)$

=> $\lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)}$

= $lim \ \frac{- 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)}{2sin(x)cos(x)}$

= $lim \ \frac{-8cos^2(x)+4sin^2(x)-1}{2cos(x)} = \frac{-8+4*0-1}{2*1} = -9/2$

Any questions?
• May 3rd 2009, 08:42 AM
manalive04
how did you calculate the cos(3x) how is that broken down?

i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

but how did u do this one
• May 3rd 2009, 08:48 AM
HallsofIvy
Quote:

Originally Posted by manalive04
how did you calculate the cos(3x) how is that broken down?

i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

but how did u do this one

Yes, $cos(2x)= cos^2(x)- sin^2(x)$ and $sin(2x)= 2sin(x)cos(x)$

Write $cos(3x)= cos(2x+ x)$ and use the sum formula: cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) so that cos(3x)= cos(2x)cos(x)- sin(2x)sin(x). Now put the formulas form cos(2x) and sin(2x) into that.
• May 3rd 2009, 08:49 AM
Rapha
Hi again.

Quote:

Originally Posted by manalive04
how did you calculate the cos(3x) how is that broken down?

i know cos(2x) is cos(x)cos(x) - sin(x)sin(x)

but how did u do this one

List of trigonometric identities - Wikipedia, the free encyclopedia

Use this formula for n=3

???

Does it help?
• May 3rd 2009, 08:52 AM
curvature
Using l'Hospital's Rule

$\lim_{x \to 0} \ \frac{cos(3x)-1}{sin^2(x)}
=\lim_{x \to 0} \ \frac{-3sin(3x)}{2sin(x)cos(x)}
=\lim_{x \to 0} \ \frac{-3sin(3x)}{sin(2x)}
=\lim_{x \to 0} \ \frac{-9cos(3x)}{2cos(2x)}
=-\frac{9}{2}
$