Solving Limits with Spherical Coordinates

[physstudent1]

Homework Statement

lim as (x,y,z) ->0,0,0 of xyz/(x^2+y^2+z^2) it says (HINT: use spherical coordinates)

Homework Equations

The Attempt at a Solution

I tried putting it in spherical coordinates but it didn't seem to help. I ended up getting pcos(theta)sin(psi)p(sin(theta)sin(psi))pcos(theta)/(p^2cos^2(theta)sin^2(psi)+p^2sin^2(psi)sin^2(theta)+p^2cos^2(psi))

I don't know what I can do with this messy thing or even why theywould tell me to use it just looks like it made it worse. By the way my teacher said we must use spherical coordinates or we will receive no credit.

 

[CoCoA]

Are you sure you don't have a better expression for your denominator using the formulas relating spherical and cartesian coordinates?

Then, if (x,y,z)->0, what is/are going to what limit in spherical?

 

[physstudent1]

oh i see you could put the denominator as p^2 correct? and then the top would go to 0 as (x,y,z) go to zero?

 

[Dick]

oh i see you could put the denominator as p^2 correct? and then the top would go to 0 as (x,y,z) go to zero?

The 'top' going to zero is not enough. The top has to go to zero faster than p^2. Does it?

 

[physstudent1]

actually wouldn't the top have a p^3 term so it would go to 0 faster then the bottom?

 

[Dick]

What's the power of p on the top?

 

[physstudent1]

it's to the third power right so you end up with a 1 in the denominator and a p in the numerator?

 

[Dick]

Yes, and p goes to zero, right? What about the trig functions? They are bounded, right?

 

[physstudent1]

yea they are bounded between -1 and 1 and p goes to 0 so the top becomes 0 and the limit is 0

 

[Dick]

Right. The quotient bounded between -p and p and p goes to zero. Well done.

 

[physstudent1]

thanks for the help