L'Hospitals Rule?

Question

Im am trying to find the limit of the function (cos x - 1) / sin x^2 as the limit approaches 0.

If you plug in 0 you get 0/0 meaning l'hospitals rule applies.

So I take the derivative of the top and bottom seperatly

I get -sin x / 2x cos x^2, which still gives me a 0 for an answer,

even if I take more derivatives I will keep getting the same answer...However I guessed in my head the answer would be -1/2, but I cant show it

Any ideas?

Answer 1

Apply L'Hopital's rule again... d/dx(-sin(x)) = - cos(x). d/dx(2xcos(x^2)) = 2 d/dx(xcos(x^2)) = 2 (x(-2xsin(x^2)) + cos(x^2)). Evaluating the top at x=0, gives -1. Evaluating the bottom at x=0 gives 2. Hence the limit is -1/2

Answer 2

You only need to apply the L'Hospital's rule twice.

lim (cos x - 1)/sin x^2, x->0
= lim -sin x/(2x cos x^2), x->0
= lim -cosx/(2 cos x^2 + 0), x->0
= -1/2, since cos 0 = 1
-----------
"0" in the next to the last step comes from the limit of 2x[cos x^2]'.

Answer 3

As you noted when you have a quotient in the indeterminant form of 0/0 L'Hospital's rule applies.

Lim x→0 of [(cos x - 1)/(sin x²)

= Lim x→0 of [-sin x / (2x cos x²)]

= Lim x→0 of [-cos x / (2cos x² - 4x² sin x²)]

= -1/2