# Homework Help: 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)?

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