# L'Hopitals Rule

• Jul 24th 2011, 01:32 AM
Arron
L'Hopitals Rule
I need to prove the following limit exists, and evaluate it

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

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

The f and g are differentiatiable, and f(0)=g(0)=0

Thus f and g satisfy the conditions of L'Hopitals Rule at 0.

Now

f'(x)=sin3x and g'(x)= 4x^3+6x

Thus by L'Hopitals Rule, the required limit exists and equals

lim
x-->0 f'(x)/g'(x) = lim x-->0 Sin3x/4x^3+6x = ?

This is where I am stuck. I know if just the basic end bit but could anyone help?
• Jul 24th 2011, 01:38 AM
chisigma
Re: L'Hopitals Rule
Quote:

Originally Posted by Arron
I need to prove the following limit exists, and evaluate it

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

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

The f and g are differentiatiable, and f(0)=g(0)=0

Thus f and g satisfy the conditions of L'Hopitals Rule at 0.

Now

f'(x)=sin3x and g'(x)= 4x^3+6x

Thus by L'Hopitals Rule, the required limit exists and equals

lim
x-->0 f'(x)/g'(x) = lim x-->0 Sin3x/4x^3+6x = ?

This is where I am stuck. I know if just the basic end bit but could anyone help?

First a little observation: if $\displaystyle f(x)= 1-\cos 3x$ then $\displaystyle f^{'}(x) = 3\ \sin 3x$...

Second: if the first application of l'Hopital rule gives an indeterminate form, then try to use l'Hopital rule again till to obtain a non indeterminate form!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 25th 2011, 01:19 PM
Arron
Re: L'Hopitals Rule
Thanks

So f"(x)= 9 cos 3x
and g"(x) = 12x^2+6

Since g"(0) = 6 which is not 0. we deduce, by combination and continuous functions, that

lim
x---> =9/6 = 3/2

Is this correct?
• Jul 25th 2011, 01:21 PM
Siron
Re: L'Hopitals Rule
Correct!