Limits of Trig Functions: How to Solve Using L'Hopital's Rule?

[DeathWish]

Homework Statement

lim(x -->0) (1-cos(14x))/(xsin(18x))

Homework Equations

None?

The Attempt at a Solution

The hint tells me to use L'hopital's rule through which I got

lim(x-->0) (sin(14x))/(18xcos(18x)+sin(18x)) (I factored out the 14 in the numerator)

That gave me a 0/0 so I did L'hopital's rule again, through which I got

lim(x-->0) (cos(14x))/(cos(18x)-9xsin(18x)) (I factored out 14/2)

This gives me 98, which is wrong. (As x-->0, the function becomes 1, leaving what I factored out as 98)

I'm not sure what to do here. Help? :(

 

[Dick]

I think it's the things you are factoring out. I don't think you factor out 14/2 to get the expression you showed at the end. Can you show how you did that? Try showing your expressions without the 'factoring out'.

 

[DeathWish]

lim(x-->0) (1-cos14x)/(xsin(18x)), this gives 0/0

L'hopital's

lim(x-->0) (14sin14x)/(sin(18x) + 18x(cos(18x))), Gives 0/0

L'hopital's

lim(x-->0) (196cos(14x))/(36cos(18x)-324xsin(18x)) This gives 49/9 which is right...

Thanks, it seems I just screwed up the calculations. Sorry for the bother

 

[Dick]

lim(x-->0) (1-cos14x)/(xsin(18x)), this gives 0/0

L'hopital's

lim(x-->0) (14sin14x)/(sin(18x) + 18x(cos(18x))), Gives 0/0

L'hopital's

lim(x-->0) (196cos(14x))/(36cos(18x)-324xsin(18x)) This gives 49/9 which is right...

Thanks, it seems I just screwed up the calculations. Sorry for the bother

You are welcome. No problem.