Newton's Method and Hoptial's Rule

1. Jul 24, 2011

redbird555

Hi all I have these problem on my test review and even after guessing one right I have no idea how one would arrive at that answer so your help is greatly appreciated.
1. The problem statement, all variables and given/known data

1.find the limit as x approaches 0 in the equation[(3/x^4)-(4/x^2)]

2.Use Newton's method to find the positive value of which satisfies . Compute enough approximations so that your answer is within .05 of the exact answer.

3. The attempt at a solution
For the limit problem I simply tried taking the derivative of the two equations and subtracting them which gave me 0/0. I know the answer to the problem is infinity but I have no idea how one would arrive at that solution?

On the second problem I really dont have a clue. I dont know how to apply that to the normal newtons method formula to a problem worded like this any help is appreciated.

2. Jul 24, 2011

HallsofIvy

Staff Emeritus
$$\frac{1}{x^4}- \frac{4}{x^2}= \frac{1- 4x^2}{x^4}$$.
Apply L'Hopital's rule to that.

"Use Newton's method to find the positive value of which satisfies ."
I have no clue what that means. There seem to be words missing after "positive value of" and "satisfies".

3. Jul 24, 2011

redbird555

Use Newton's method to find the positive value of x which satisfies x=1.5cos(x) . Compute enough approximations so that your answer is within .05 of the exact answer. (You may use any starting point you deem appropriate.)

4. Jul 24, 2011

redbird555

The only thing is my problem is 3/(x^4) instead of 1/(x^4) I tried to apply hopitals rule to your example and I came out with just zero after taking the derivative of the result and then plugging in zero back into the derivative.

5. Jul 24, 2011

SammyS

Staff Emeritus
@ redbird555,

Does L'Hopital's rule apply to $\displaystyle \lim_{x\to 0}\,\frac{3- 4x^2}{x^4}\ ?$ Why not?

You should be able to do this without L'Hopital .

6. Jul 24, 2011

redbird555

It would and your right it wouldnt even really need to apply hereI could just plug in zero. However if I do that I get a divided by 0 answer and I know the answer to the problem is a negative infinity so I'm not sure whats up.

7. Jul 24, 2011

SammyS

Staff Emeritus
Looks like +∞ to me.

8. Jul 24, 2011

redbird555

ugghhh sorry I had a brain fart for a second it is positive infinity. thanks for all the help guys does anyone know how to complete the 2nd one I'm completely lost?? I relisted the problem with the correct info.

9. Jul 24, 2011

SammyS

Staff Emeritus
Do you know Newton's method ?

To get a starting value, graph:
y=x

and

y=cos(x)​

On the same set of axes & see approx where they intersect.

10. Jul 24, 2011

redbird555

yes i understand how newtons method works but I think the y=1.5cosx is confusing me.
I should be able to take any value for x and plug it in so lets say I take 3. Now that per newtons method 3-[1.5cos(3)/(-3sin(3)/2)] this gives me -4.01525 and that is not correct which is why I'm stumped

Last edited: Jul 24, 2011
11. Jul 24, 2011

SammyS

Staff Emeritus
x = 3 is a poor choice.

See the graph I suggested.

What are you using for your function?

What's the derivative?

Attached Files:

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12. Jul 24, 2011

redbird555

ok lol im either very confused or there was some mis-communication. I'm using 1.5cos(x) as the function. so by that graph 1 would look like a logical choice therefor newtons method reads x1-[f(x)/f'(x)] so I would get 1-[1.5cos(1)/(3sin(1)/2)] correct? If so that leaves me with a final value of 1.64209 which is not correct either. I've done quite a few newtons method problems on this hw and was able to do them fine thats why I keep thinking its some organization mistake in the equation.

13. Jul 24, 2011

SammyS

Staff Emeritus
From one of your earlier posts:
I thought you were solving x = 1.5 cos(x) .

14. Jul 24, 2011

redbird555

Well its asking for an x value that would satisfy that equation and as far as I could gather thats what I was trying to solve for in my calculations? I just need to figure it out before 1130 seeing as thats when my hw is due and this is the last problem I have

15. Jul 24, 2011

SammyS

Staff Emeritus
Newton's method finds a root of a function. You need a function, f(x), such that f(x) = 0 for the value of x that solves your equation. such a function is:

f(x) = x - 1.5cos(x)

Find f ' (x) for that function.

Then $\displaystyle x_1 = x_0-\frac{f(x_0)}{f'(x_0)}\,.$

In general: $\displaystyle x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}\,.$