Question

lim cos2x - cos 3x/cos4x - 1 x--> 0

Answer 1

Just to be sure, is it this?
\[\lim_{x \rightarrow 0}\frac{\cos 2x - \cos 3x}{\cos 4x - 1}\]

Answer 2

that's the same thing I got also

Answer 3

http://www.wolframalpha.com/input/?i=limit+cos2x+-+cos+3x%2fcos4x+-+1+x--%3e+0&incTime=true

Answer 4

yup, I need to evalute the limit

Answer 5

The first thing you should try is direct substitution: Plugging in 0 for x.

Answer 6

eh.. it direct submission wouldn't make everything zero right?

Answer 7

What do you mean? :3
Plug in 0 for x and evaluate. What do you get?

Answer 8

@iPwnBunnies Plug in 0 for x, we'll get \(\frac{0}{0}\).

Answer 9

if you have \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), where are indeterminate, you can use l'Hopital's rule.
\[\lim_{x\rightarrow0}\frac{\cos(2x)-\cos(3x)}{\cos(4x)-1}\]\[=\lim_{x\rightarrow0}\frac{[\cos(2x)-\cos(3x)]'}{[\cos(4x)-1]'}\]\[=...\]That ' is differentiation.
In this question, you need to apply this rule twice.