tough(?) limit of trig functions - no Hopital's allowed.can you do it?

Question

this is the limit (fairly easy if done with l'Hopital's rule) lim when x --> 0 of (1-cos5x)/(cos7x-1).
Thank you,
MM

Answer 1

1-cos5x=1-(cos^2(5x/2)-sin^2(5x/2))
=1-1-2sin^25x/2=-2sin^2(5x/2)
similarly
1-cos7x=-2sin^2(7x/2)
so limit x-->0 (1-cos5x)/(1-cos7x)
=lim x-->0-2sin^2(5x/2)/-2sin^2(7x)/2)
=lim x--->0 [sin(5x/2)/(5x/2)]^2/[sin(7x/2)/(7x/2)^2]
[(5x/2)^2/((7x/2)^2]
=25/49 answer

Answer 2

lim x--> -2(fx) ??? this notation does no longer make any experience of course this cut back returned does no longer exist by utilising ability of plugging in -2 because of the fact the denominator could be 0. what you're arranged to do, is element x^2 out of the numerator, cancel the loosen up element with the denominator, and you gets 4 as quickly as you plug in -2 B could be a piecewise function utilising the unique and a 2nd equation f(-2) = 4 to plug the hollow. subsequently, the needs are precisely the comparable (for particular) different than you on the instant filled indoors the hollow.

Answer 3

I think the way to do this is to use Trig identities to modify the numerator and denominator so that you can cancel things out.

There is an identity for cos(Kx) that you can try.

But you might have more luck multiplying the top and the bottom by cos(7x) + 1. That will allow you to change the bottom into a Sin.