A limit question stuck.

[life is maths]

Homework Statement

Hi, my question is
lim$_{x\rightarrow0}$x²$\frac{cos(cot(x))}{sinx}$

Homework Equations

The Attempt at a Solution

I thought maybe I could make $\frac{cos(cot(x))}{sinx}$ similar to $\frac{sinx}{x}$, but couldn't find a proper way for it. Dividing sinx by cosx and multiplying does not take me anywhere, neither does trying to manipulate cotx. Is there something I need to see but cannot see?
L'Hospital and derivatives are not valid solutions since we haven't learned them yet.Could you please help me find a way? Thanks for any help.

 

[dynamicsolo]

It will help if you factor the function's expression first to break this into limits you do know how to work with:

$$( \lim_{x \rightarrow 0} \frac{x}{\sin x} ) \cdot ( \lim_{x \rightarrow 0} x \cos (\cot x) )$$

For the second limit, you will need to consider how cotangent behaves and what the cosine of that value is if it were treated as an angle. (Plot cos(cot x) to check on this.) As x approaches zero, it behaves in a crazy way, but one which you may have seen before. What happens when you multiply that by x ? What limit method do you know for dealing with something like that?

 

[life is maths]

Thanks a lot dynamicsolo.

The first part of the limit goes to 1, right? As for lim$_{x\rightarrow0}$(cos(cot(x)), since cosx is always between -1 and 1, and since x goes to zero, when I multiply them, I get 0. Is it true?

 

[lurflurf]

^Yes that is true.

 

[dynamicsolo]

Thanks a lot dynamicsolo.

The first part of the limit goes to 1, right? As for lim$_{x\rightarrow0}$(cos(cot(x)), since cosx is always between -1 and 1, and since x goes to zero, when I multiply them, I get 0. Is it true?

Yes, you get to use a trigonometric limit and the "Squeeze Theorem" in the same problem!