Thanks I'll try this.

Also I just want to clarify that when you want to find the limit of a function with 1 variable you manipulate the function and then just substitute the value.

But when you have a function of 2 variables, you can't just substitute in the value can you?

Reply 3

Indeterminate

Original post by Vorsah

Informally, if I wanted to find the limit, as

(x,y) \rightarrow (0,0)

of, say,

f(x,y) = x^2 + y^2

then I could see that this limit is zero since both x AND y are approaching zero.

You could prove this rigorously by using the definition of a limit on a vector (x,y).

However, you CANNOT substitute in if you've got something that'll give something like 0/0.

(edited 9 years ago)

Reply 4

pseudonymegg

limits why couldn't I remember limits?
that's the day Aaron got his haircut... man he looked so cute
FOCUS
what was on the board behind Mrs Norberry's head?
If the limit doesn't approach anything, the limit does not exist
THE LIMIT DOES NOT EXIST!!

Reply 5

Indeterminate

Original post by pseudonymegg

That's false. The limit does exist in this case.

Reply 6

Vorsah

Original post by Indeterminate

Informally, if I wanted to find the limit, as

(x,y) \rightarrow (0,0)

of, say,

f(x,y) = x^2 + y^2

What do you when you get 0/0 if you can't substitute in?
Do you have to make one of the variables constant and show the limit does not exist?

Reply 7

Indeterminate

Original post by Vorsah

What do you when you get 0/0 if you can't substitute in?
Do you have to make one of the variables constant and show the limit does not exist?

Well, the easiest way to show that a limit of a function in 2 variables doesn't exist is if the directional limits are different. In other words, you set x and y, in turn, equal to zero and see what the limit from those respective directions is. If they're different, then it doesn't exist.

If the limit exists but substitution gives 0/0, then the most obvious way is to bound it by something you already know about and then use the sandwich rule.

Reply 8

Vorsah

Original post by Indeterminate

I need help with q6. I have done the first part, but i need help with ' but that there its now way to define f at the origin such that f is continuous at (0,0)'

image.jpg224.9KB

Reply 9

Indeterminate

Original post by Vorsah

I need help with q6. I have done the first part, but i need help with ' but that there its now way to define f at the origin such that f is continuous at (0,0)'

Hint:

For the function to have any chance of being continuous there, it must take the value of its limit at that point. For, if it didn't, then there would be a "jump" and it would be discontinuous.

Reply 10

Vorsah

So I should show the limit doesn't exist

So can I substitute in x=mx and y=mx^2 and show the limit doesn't exist?

Edit:

I subbed in x=y^2 and y=y and got 1/0 so have I correctly shown the limit doesn't exist?

(edited 9 years ago)

Reply 11

Vorsah

Original post by Indeterminate

Hint:

For the function to have any chance of being continuous there, it must take the value of its limit at that point. For, if it didn't, then there would be a "jump" and it would be discontinuous.

I have another question. Evaluate the limit I the the function f(x,y)= y^3/(x^2+y^2) as f(x,y)---->(0,0)

When you substitute you get 0/0. So how would you evaluate the limit? If I take the directional limits, like you said, I get y (when x=0) and 0 (when y=0) so does that limit not exist?

If the question was prove the limit = 0 then I would know that I have to use the definition of a limit to prove this.

(edited 9 years ago)

Reply 12

Indeterminate

Original post by Vorsah

No, the limit does exist. What you're meant to be doing is showing that it's not continuous at the origin.

Original post by Vorsah

The directional limit approach doesn't work here since you get a variable y, which could be 0.

The limit exists and it is 0. Try bounding f by some other function whose limit at the origin is 0 and use the sandwich rule.

Reply 13

Vorsah

Original post by Indeterminate

No, the limit does exist. What you're meant to be doing is showing that it's not continuous at the origin.

The directional limit approach doesn't work here since you get a variable y, which could be 0.

The limit exists and it is 0. Try bounding f by some other function whose limit at the origin is 0 and use the sandwich rule.

I don't understand how you get the limit as 0. Do you have to use the epsilon/delta definition of a limit?

Reply 14

Indeterminate

Original post by Vorsah

I don't understand how you get the limit as 0. Do you have to use the epsilon/delta definition of a limit?

First define a vector X = (x,y) and consider when it's close to the origin, ie the case where

|X| < 1

. Then, importantly,

x^2 + y^2 < 1

and you can go on to use your definition of a limit to make an appropriate choice of delta.

EDIT: even though it says "evaluate" and not "prove", there's nothing better than making use of a definition to justify your answer.