Question

Find \[ \lim_{x\to 0}\frac{cos(ax)-cos(bx)}{x^{2}}} \]

Answer 1

can u use directly this limit formula:
\[\lim_{x \rightarrow 0}\frac{ 1-\cos cx }{ x^2 } = c^2/2\]

Answer 2

\[ \lim_{x -> 0} \frac{cos(ax)-cos(bx)}{x^{2}} \]

Answer 3

I don't think so since I have never see this formula before.

Answer 4

what about L'Hopital's ?

Answer 5

I've been told that I should use L'Hopital's Rule, but I don't know how to do this.

Answer 6

Yeah, can you show me how?

Answer 7

Or should I show you my work?

Answer 8

show your work,
did u differentiate the numerator ?

Answer 9

Yeah , well I got this for numerator: \[ x(sin (ax) - sin (bx)) \]I don't know what it is but I think I did something wrong.

Answer 10

Hmm not sure where the x is coming from <:o

Answer 11

when u differentiate with respect to x, a is constant.
so,
d/dx (cos ax) = -sin (ax) d/dx(ax) <----chain rule.
aware of chain rule ?

Answer 12

\[\frac{ d }{ dx } \cos (ax) - \cos (bx) = \ b sin (bx) - a sin (ax)\]Right?

Answer 13

absolutely!
and d/dx(denominator) =?

Answer 14

2x, lol.

Answer 15

now check whether the form is still 0/0 ??

Answer 16

So \[\lim_{x \rightarrow 0} \frac{\cos(ax)-\cos(bx)}{x^{2}} = \lim_{x \rightarrow 0} \frac{b \sin (bx) - a \sin (ax)}{2x}\]It is still 0/0 so I have to use L'Hopital's rule again, right?

Answer 17

I think I got it.

Answer 18

right!
tell me what final answer u get....

Answer 19

\[\frac{ b^2 - a^2 }{ 2 }\]Thanks. Sorry about that. I just got confused with al of those things, lol.

Answer 20

thats correct.
welcome ^_^