# Prove a limit exists and evaluate it

1. Aug 24, 2009

### 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'Hopitals 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

2. Aug 24, 2009

### 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.

3. Aug 27, 2009

### flatmaster

You could use the power series of the cosine.

4. Aug 28, 2009

### uart

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.

5. Aug 28, 2009

### james.farrow

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