Limiting Values of 2cos^2(x)/cos(3x) at x->pi/2

[Dell]

lim
x->pi/2 (1+cos2x)/cos3x

i tried all the identities and got

2cos^2(x)/cos(3x)

what now? i used cos(a+b)=cos(a)*cos(b)-sin(a)sin(b)
but no help there

 

[Dick]

I don't think you tried 'all of the identities'. The denominator hasn't changed at all. What can you do to it?

 

[Dell]

the best i could come up with is

lim
x->pi/2 (1+cos2x)/cos3x

lim
x->0 2cos^2(pi/2-x)/cos(3(pi/2-2))

lim
x->0 (2sin^2(x))/sin(3x)

lim
x->0 2sin(x)*-1/3

=0

 

[lanedance]

can you use L'hopitals? good to check with anyway...

 

[lanedance]

EDIT: as below

in the mean time its worth trying Dick's suggestion though...

 

[Dick]

the best i could come up with is

lim
x->pi/2 (1+cos2x)/cos3x

lim
x->0 2cos^2(pi/2-x)/cos(3(pi/2-2))

lim
x->0 (2sin^2(x))/sin(3x)

lim
x->0 2sin(x)*1/3

=0

Ok, you can do it that way. cos(3(pi/2-x))=(-sin(3x)), though.

 

[Dick]

With last comment in mind, I'm trying to pick the flaw in you logic... i think you miss a negative when you subsitute x->pi/2-x, and can't just cancel the (2sin^2(x))/sin(3x) to 2sin(x)*1/3.

This is beacasue of the trig identities allow you to write sin^2x as sin2x and so on, think what thismeans in terms of teh taylor series...

in the mean time its worth trying Dick's suggestion though...

I guess I was assuming the OP knew lim sin(x)/sin(3x)=(1/3) and didn't think sin(x)/sin(3x)=(1/3). Hope I wasn't wrong.

 

[lanedance]

nope, my bad, made a quick arithmetic error, all looks good