Limits x approach 0, show that (cos2x-cos3x)/(cos4x-1)= -5/16

Question

pl provide the answer this question

Answer 1

Given lim(x-->0) [(Cos2x-Cos3x)/(Cos4x-1)]

Here it is in the form 0/0,So apply L - Hospital rule.

= lim(x-->0)[(-2Sin2x+3Sin3x)/-4Sin4x]

Again apply L- Hosptal rule.

= lim(x-->0) [(-4Cos2x+9Cos3x)/-16Cos4x]

Evaluting limit

= [-4(1)+9(1)]/-16(1)

= (-4+9)/-16

= -5/16