# Multivariable calculus: Limits (2)

1. Sep 16, 2007

### kingwinner

Note: <= means less than or equal to, >= means greater than or equal to

1) Prove formally that
lim xy^2 / (x^2 + y^2) = 0
(x,y)->(0,0)
without breaking the vectors into components.

Let X=(x,y). We want to prove that for all epsilon>0, there exists a delta>0 such that if 0<||X-0||=sqrt(x^2 + y^2)<delta, then |xy^2 / (x^2 + y^2)|<epsilon

Starting with |xy^2 / (x^2 + y^2)| <= ? For here HOW can I find an appropriate inequality that can help me find a delta that works? Remember that here we are not allowed to break the vector into individual components.

2) Evaluate lim e^(x+z) / {z^2 + cos[sqrt(xy)]}.
(x,y,z)->(1,0,-1)

Solution:
We must have xy>=0
By continuity, the limit equals (direct substitution)
e^(1-1) / {(-1)^2 + cos[sqrt(0)]} = 1/2

I have no idea why they can use continuity/direct subtitution to eavluate the limit...
I believe that as (x,y,z) approachs (1,0,-1) from ALL direction, the limit must exist and equal , but in the case, say, I approach (1,0,-1) through the route (0.99, -0.01, -0.99) ->(0.999, -0.001, -0.999) is clearly impossible because xy=(0.999)(-0.011)<0 which means it is out of the domain. Then how can the limit still exist?
As far as I know, at least some neighbourhood of (1,0,-1) has to be defined for the possibility of a limit to exist...

I am still terribly confused no matter how many times I read the definitions and theorems from my textbook. Any help or explanation will be greatly appreciated!

Last edited: Sep 17, 2007
2. Sep 16, 2007

### EnumaElish

1) How do you write out ||X-0||?
2) Why worry about a sequence that is not in the domain? The point is, the sequence you use must lie within the domain, and as long as it does, you have a limit (in this case).

3. Sep 17, 2007

### kingwinner

1) ||X-0|| = sqrt (x^2 + y^2), but how does this help?

2) For example,
lim sqrt x doesn't exist since the left hand limit doesn't exist.
x->0
For the limit to exist, the left and right hand limits must exist and must be equal.

So similarly, in this case, the limit must exist as (x,y,z)->(1,0,-1) from every direction, but in some directions the limit clearly doesn't exist because the function isn't even defined. Having said these, I now don't understand why it's still possible for the limit to exist and equal to 1/2...

4. Sep 17, 2007

### CompuChip

since $x^2 \le x^2 + y^2$, certainly $|x| \le \sqrt{x^2 + y^2} = ||(x, y)||$ and similarly $|y| \le ||(x, y)||$.
Now
$$| x y^2 / (x^2 + y^2) | = |x| |y|^2 / ||x||^2$$.
Use the above inequalities to estimate |x| and |y| and (assuming that X is not 0, divide out some norms of X), and you will be done.

5. Sep 17, 2007

### HallsofIvy

Staff Emeritus
Put
$$\frac{xy^2}{x^2+ y^2}$$
into polar coordinates. That way, "(x,y) going to (0,0)" just means that r goes to 0 irrespective of $\theta$.

The definition of "continuity" is "$\lim_{x\rightarrow a} f(x)= f(a)$"!

From ALL possible directions. From all directions in the domain.

6. Sep 20, 2007

### kingwinner

For example,
lim sqrt x doesn't exist since the left hand limit doesn't exist.
x->0
For the limit to exist, the left and right hand limits must exist and must be equal. It is NOT just from all directions IN the domain...

So similarly, in this case, the limit must exist as (x,y,z)->(1,0,-1) from EVERY direction I believe, but in some directions the limit clearly doesn't exist because the function isn't even defined. Having said these, I now don't understand why it's still possible for the limit to exist and equal to 1/2...