Finding limit help.

1. Oct 17, 2004

KataKoniK

Hi,

I was wondering if anyone here can help me with the following question.

lim 1 - cos(x³ sin x) / x^6 sin² x
x->0

I have tried multiplying the top and the bottom by 1 + cos(x³ sin x), so I can get

(sin²(x³ sin x) / x^6 sin² x) * 1 / 1 + cos(x³ sin x)

I get stuck there. I don't know what to do next. Any help would be great thanks.

2. Oct 17, 2004

arildno

Welcom to PF!
Just to make sure I understand your notation, you are seeking:
$$\lim_{x\to0}\frac{\sin^{2}(x^{3}\sin(x))}{x^{6}\sin^{2}(x)}\frac{1}{1+\cos(x^{3}\sin(x))}$$

Use the sneaky $$\sin(y)\approx{y},\cos(y)\approx1 (y\approx0)$$ to find the limit.
You should end up with $$\frac{1}{2}$$

3. Oct 17, 2004

singleton

sin^2(x^3sin(x)) / (x^3sin(x)) * 1 / (x^3sin(x)) * 1 / 1 + cos(x^3sin(x))

The limit of the first one is 1 (since the argument of the sin and denominator are the same) and the limit of the last is 1/2 (cos0 is 1 so that is 1/2)

that is all about I know ^^

The middle part poses a problem for my limited experience ><

4. Oct 17, 2004

KataKoniK

Thanks for the reply. I am currently taking first-year Calculus in University. It is very difficult.

I do not understand what the $$\sin(y)\approx{y},\cos(y)\approx1 (y\approx0)$$ notation is. I have never seen it before, unless I am missing something.

5. Oct 17, 2004

KataKoniK

Nice eye singleton. I didn't even notice the part of how I could have broken up x^6 sin x

Thanks for the help. I'll try it out.

6. Oct 17, 2004

arildno

What it means, is that when the argument of sine (y) is approximately zero, then
sin(y) is approximately equal to y.
You've seen the following limit before:
$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$ (Right?)
This implies that $$\sin(x)\approx{x}$$ when x is close to zero..

7. Oct 17, 2004

singleton

nevermind I'm wrong! (can't seem to get rid of an undefined anyway I try and cut it) try to google the method the other poster suggested!

Last edited: Oct 17, 2004
8. Oct 17, 2004