# Multivariable calculus: Limits (2)

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:

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EnumaElish
Homework Helper
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).

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...

CompuChip
Homework Helper
We want to prove that for all epsilon>0, there exists a delta>0 such that if ||(x, y)||<delta, then |xy^2 / (x^2 + y^2)|<epsilon
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.

HallsofIvy
Homework Helper
kingwinner said:
1) Prove formally that
lim xy^2 / (x^2 + y^2) = 0
(x,y)->(0,0)
without breaking the vectors into components.
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 $\theta$.

kingwinner said:
I have no idea why they can use continuity/direct subtitution to eavluate the limit...
The definition of "continuity" is "$\lim_{x\rightarrow a} f(x)= f(a)$"!

kingwinner said:
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?
From ALL possible directions. From all directions in the domain.

444
From ALL possible directions. From all directions in the domain.

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...