lim[x approaches 0, (1-cos5x)/(7x^2)]?
Could you explain this in exact detail of how to get the answer. I know the answer is 25/14, but I am unsure of how it comes to be that.
I was told it could be done using L'hopitals rule but i really don't understand. Any detailed help would be greatly appreciated!
Update:Thanks so much you guys for all of your answers! everything makes so much more sense now! does anyone know how do i give out points to all of you? all answers were great! and fast!
Comments
When you use l'hopital's rule, you first need to make sure you get an indeterminate form. If that's the case (which it is here), take the derivative of the top and the derivative of the bottom. If you still get an indeterminate form, do it again.
lim x->0 [(1-cos5x)/(7x^2)]
lim x->0 [(5sin5x) / (14x)]
You get 0/0 again, so use l'hopital's rule once again.
lim x->0 [(5sin5x) / (14x)]
lim x->0 [(25cos5x) / (14)]
= (25cos0) / (14)
= 25 / 14
substitute x=0, we have 0/0
Apply L'Hopital's rule
lim x-->0 (1-cos 5x) / 7x^2
= lim x-->0 5 sin 5x / 14 x
= lim x-->0 25 cos 5x / 14
= 25/14
differentiate the numerator and denominator
).
(lim -> 0) 5sin(5x) / 14x = 25cos(5x) /14 = 25/14