Prove a limit exists and evaluate it

[james.farrow]

We are required to find

lim x-> 0 1- cos(2x)/x^3 + 4x^2

Let f(x) = 1 - cos(2x) & g(x) = x^3 + 4x^2

f'(x) = 2sin(2x) so f'(0)=0
g'(x) = 3x^2 + 8x & g'(0) = 0

so by l'hospital's rule

lim x -> 0 2sin2x/3x^2 + 8x if it exists

Now

f''(x) = 2(2cos2x) & g''(x) = 6x + 8 = 2(3x + 4)

But f''(0) = 4 and g''(0) = 8

Shouldn't f''(0) = g''(0) = 0 here?

I'm a little bit stuck now?

or is the limit just f''(0)/g''(0) which is 4/8 which is a half?

Many thanks

James

 

[Дьявол]

The limit is f''(0)/g''(0), and it is 4/8=1/2.

If f"(0)/g''(0)=0/0, then the limit will go to infinity, hence no limit will exist.

Regards.

 

[flatmaster]

You could use the power series of the cosine.

 

[uart]

If f"(0)/g''(0)=0/0, then the limit will go to infinity, hence no limit will exist.

Regards.

BTW that is simply not true. You can keep applying L'Hoptials rule as many times as you like, while ever you are faced with 0/0. Take for example x^4 / ( 1-cos x )^2, you'd need to go to the 4th derivative in this case before L'Hopitial's finally cracks it.

 

[james.farrow]

Cheers everyone I think I'm getting the hang of it...