1. Apostol limit problem

I can't afford the Apostol calculus vol. 2 there's a printing mistake in my copy of Apostol and I'm not sure how to prove this:

let f(x,y)={xsin(1/y) if y doesn't equal 0
and f=0 if y=0
prove that the iterated limits are not equal and that the f(x,y)->0 as (x,y)->(0,0)

2. Re: Apostol limit problem

If y is fixed (not equal to zero, of course), then the limit as x goes to zero is zero.

If x is fixed, then the limit as y goes to zero doesn't exist.

As a result, the two-variable limit does not exist - two different paths give two different results.

- Hollywood

3. Re: Apostol limit problem

thanks, so by holding x, y fixed you can prove the limit when (x,y)->0. is there a text that covers this please?
can I somehow use the equation |xsin y-1|<=|x|<=|sqrt (x^2+y^2)|

4. Re: Apostol limit problem

You can prove the limit doesn't exist - if the limit is different depending on which path you take to the origin - but you can not prove that the limit does exist. For that, you can use the squeeze theorem (or its two-dimensional equivalent), which is what you seem to be trying to do. You can also convert to polar coordinates, so the condition "(x,y) goes to 0" converts to the one-dimensional "r goes to 0", which we know how to deal with.

- Hollywood

5. Re: Apostol limit problem

so if you prove the limit of the outer terms is equal then by Squeeze Theorem the center limit is also the same. if you take the distance function (x^2+y^2+z^2)^(1/2) then as this goes to zero you approach the origin from all directions?

6. Re: Apostol limit problem

Yes, if $a(x,y) \le b(x,y) \le c(x,y)\,$ and $\lim_{(x,y)\to{0}}a(x,y)$ and $\lim_{(x,y)\to{0}}c(x,y)$ both exist and are the same value L, then $\lim_{(x,y)\to{0}}b(x,y)=L$.

And if you convert to - in your case, it would be spherical coordinates $(\rho,\phi,\theta)$ - if the limit as $\rho$ goes to 0 exists and does not depend on $\phi$ or $\theta$, then the limit as (x,y,z) goes to zero also exists and is that same number.

- Hollywood