L'hopital's rule?

Question

Use l'hopitals rule to sokve the following

lim ((cos(x) - 1) / x^2)
x->0

Answer 1

Read my other answer for the explanation for this one. :-)

You have zero on top and zero on bottom, so you can certainly go ahead and use L'hopitals.

Derivative of top and bottom.

Cos(x) -1 becomes (-sinx)
and x^2 becomes 2x.

Now you have -sinx / 2x which you need to evaluate for x->0. On top you have sin(0) which is zero, and on bottom 2(0) is zero. You have 0/0 again, so L'hopital again.

-Sin x becomes -Cos x
2x becomes 2.

Now you get -Cosx / 2, evaluated for zero that's -1/2.

Hope that helped,
UMRmathmajor

Answer 2

lim ((cos(x) - 1) / x^2) =
x->0
lim (-sin(x))/ 2x) =
x->0
lim (-cos(x) / 2) = -1/2
x->0