Finding Limit Help: Get Expert Assistance

[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.

 

[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 $$\frac{1}{2}$$

 

[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 ><

 

[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.

 

[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.

 

[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..

 

[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!

 

[KataKoniK]

Thanks for the help you two!