Limit of a function

1. Oct 1, 2011

athrun200

1. The problem statement, all variables and given/known data
See the first photo.Q45

Can I use the method similar to 43?
I don't understand what does "condering (x,0,0)or (0,y,0) mean.

2. Relevant equations

3. The attempt at a solution
see photo 2.
I only know how to do 43
I have no idea for 45

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2. Oct 1, 2011

vela

Staff Emeritus
For a limit to exist, you can't get different answers depending on the direction you approach a point. When the problem says to consider points of the form (x,0,0), it's asking you to calculate the limit as you approach the origin along the x-axis.

3. Oct 1, 2011

SammyS

Staff Emeritus
BTW: What do you conclude about #43?

4. Oct 1, 2011

flyingpig

For 45 try this

$$f(x,y,z) = \frac{x + y + z}{x^2 + y^2 + z^2}$$

What happens when you cut with x = x and with y,z =0? What happens to f?

Last edited: Oct 1, 2011
5. Oct 2, 2011

athrun200

It becomes $\frac{1}{x}$
Then, as x$\rightarrow$0, $f(x,y,z)$ doesn't exist?

6. Oct 2, 2011

athrun200

It seems if we have denominator, then the function is not continuous at origin.

7. Oct 2, 2011

HallsofIvy

??? The problem does not ask about continuity, only about the limit at the point. It is possible for the denominator of a fraction to go to 0 and the limit still exist at that point.

In problem 43,
$$\lim_{(x,y)\to (0,0)}\frac{x^2- y^2}{x^2+ y^2}$$
if we take y= mx, as suggested, we get
$$\frac{x^2- m^2x^2}{x^2+m^2x^2}= \frac{x^2(1- m^2)}{x^2(1+ m^2)}= \frac{1- m^2}{1+ m^2}$$
as long as x is not 0.

If the limit exists, then the limit of that should be the same for all m. Is it?

8. Oct 2, 2011

athrun200

It seems we use the similar method to prove continunity. If not, how do I prove that problem 43 continuous at (0,0)?