# What is the mean of that in a find limit question !

1. Dec 29, 2008

what is the mean of that in a "find limit" question !

"find limit" of function questions, i see it in two forms, i want the difference in meaning and also in the solution method for each, in details please.

First:
question say that the function $$f(x,y) = x^2 + y$$
and want me to find the limit of function [text]f(x,y)[/tex] when $$(x,y) \rightarrow (0,0)$$
also want me to see if it is continous, and the detailed prove for the two previous requests.

Second:
question say that the function $$f(x,y)$$ has two values,
1) $$f(x,y) = x^2 + y$$ at $$(x,y)\neq(0,0)$$
2) $$f(x,y) = x$$ at $$(x,y)=(0,0)$$

as mentioned above, i want the meaning and solution method ( get limit and check continous. ) and proof. for the two cases.

waiting for replies ..

2. Dec 29, 2008

### NoMoreExams

Re: what is the mean of that in a "find limit" question !

What have you tried so far, you have to show a little of your thought process.

3. Dec 29, 2008

### HallsofIvy

Staff Emeritus
Re: what is the mean of that in a "find limit" question !

The meaning is exactly the same in both cases and the "method" varies only in the definition of the function f. Sorry to be so short but if you are dealing with functions of two variables, you should be well familiar with the definition of "limit" of a function of one variable. If you will at least give what you think it to be, we will try to help you with two variables. (It's a straightforward extension.)

4. Dec 29, 2008

Re: what is the mean of that in a "find limit" question !

OK, my reply can go into the following points:

1- i should check that the value of the limit doesn't change by changing the curve, for example, if we calculated the limit on the curve y=mx and y = x, the values of the limit on the two curves should be constant, but in that case i saw in some problem when the result of the limit on the first-trying curve like y=mx, when the result contains m, the answer be that the limit not exist ! and i don't why !.

2- about the second case, ( two values for the function ), i don't know what is the use of the second value ? in the answers of the problem ( in a book ) i saw that he use only the first value ( when (x,y) not equal (0,0) ), so why that ? and what is the meaning and use of the second value for function f ?.

3- in some problems that from the second case i mentioned in the main topic, when i put y=mx and then x=0 for example, i get zero at the bottom, for example:
(1/0 + m), what should the answer here be ? and how i write the result at final ?

i may have more questions, but that's enough for now ..

5. Dec 29, 2008

### Tac-Tics

Re: what is the mean of that in a "find limit" question !

It looks like you miscopied the problem in the second part. In the second half of the definition, it says f(x, y) = x when (x,y) = (0,0). But this is the same as saying f(0, 0) = 0, which is what you would get from the first half of the definition. So, again, I think you have the problem written incorrectly.

Furthermore, what class is this for? Talking about limit and multi-parameter functions, I'd guess multi-variable calculus. But talk about "y = mx" makes it look like you're not quite that far yet.

6. Dec 29, 2008

Re: what is the mean of that in a "find limit" question !

yes, that's in calculus, limits of multi-variables functions, and i studied that time ago but found that i'm till now can only solve problems but didn't understand how that works in general way.

and about the second part, it's correct, and another example the second part in it is f(x,y) = 0 when (x,y)=(0,0) !

So, i don't know why the second part is here now.

ok, in general, what is the use of dividing the function into two parts and limiting every part with condition like (x,y)=(0,0) or (x,y)=/=(0,0) ? means how that will affect on the way we solve the problem ?

7. Dec 31, 2008

Re: what is the mean of that in a "find limit" question !

Any Help !

8. Dec 31, 2008

### HallsofIvy

Staff Emeritus
Re: what is the mean of that in a "find limit" question !

Because saying "$\lim_{x\rightarrow (a,b)} f(x,y)= L$" means that if (x,y) is "close enough" to (a,b), then f(x,y) is "close" to L (where "close" is that $\delta$ $\epsilon$ part of the limit definition). If it were possible to reach two different limits by approaching along two different paths, there would be points "close" to (a,b) giving two different numbers that cannot both be (arbitrarily) "close" to L.

However, you this is not a good way to prove that a limit does exist because you cannot check all paths! (You can check all straight lines, y= mx, but that is not sufficient. There exist functions where the limit along any straight line is the same but the limit along a parabola is different.)

I would recommend using some limit theorems if you can. In particular:
$\lim_{(x,y)\rightarrow (a,b)} [f(x,y)+ g(x,y)]= \lim_{(x,y)\rightarrow (a,b)} f(x,y)+ \lim_{(x,y)\rightarrow (a,b)} g(x,y)$
and
$\lim{(x,y)\rightarrow (a,b)} f(x)g(y)= (\lim_{x\rightarrow a} f(x))(\lim_{y\rightarrow b} g(y))$

There are not "two values", there are two "formulas" applicable for different values of (x,y). But the last one "f(x,y)= x if (x,y)= (0,0)" does not make sense- why say "= x" when x is required to be 0? It might be "f(x,y)= 0 if (x,y)= (0,0)" or "f(0,0)= 0". Since the value at (0,0) is irrelevant to the limit there, that wouldn't really make sense either. However, "f(x,y)= x if (x,y)= (x,0)" would make an interesting problem. In that case, I would recommend considering, first, f(x,y)= x2+ y for all (x,y), determining that limit as x goes to 0, then f(x,0)= 0, determing the limit as x goes to 0 (y is already 0). If those are the same, then that is not the limit. If they are different, there is no limit.

I have no idea what you mean in (3). Can you give an example?

9. Dec 31, 2008

### tiny-tim

You sometimes get functions which are defined in one way everywhere except, say, the line y = 0, and another way on that line, but with the same value at (0,0).

Then it will be discontinuous everywhere except (0,0), and it may or may not be continuous at (0,0).

But you'd need a whole line, not just one point, to be defined differently, so I don't understand the point in this question either.

10. Dec 31, 2008

Re: what is the mean of that in a "find limit" question !

So, Can i check it on $$y=mx$$ and $$y=x^2$$ for example, is that enough or also can not be sufficient ?

i think you mean -> "If those are the same, then that is the limit" NOT ", then that is not the limit", Right ?

i mean, for example:
f(x,y) = y/x ; for (x,y) =/= (0,0)
Can you check for limit existence for that function please?, and you will know what i mean.

11. Dec 31, 2008

Re: what is the mean of that in a "find limit" question !

it's clear now, sorry.

12. Dec 31, 2008

Re: what is the mean of that in a "find limit" question !

is my solution for the following example now is correct ?:

ex: For the following function:
$$f(x,y) = \frac{1}{y} sin(xy) ; y \neq 0$$
And, $$f(x,y) = x ; y = 0$$
Find the following:
1- limit when x->0 of ( limit when y->0 of ( f(x,y) ) )
2- limit when y->0 of ( limit when x->0 of ( f(x,y) ) )
3- limit when x->0 & y->0 of ( f(x,y) )

Solution for (1):
first, limit when y->0 of f(x,y); would be written as limit when y->0 of x; because f(x,0) equals x
as, limit when y->0 of x equals x
then, limit when x->0 of x equals 0
so, result would be ZERO

Solution for (2):
first, limit when x->0 of f(x,y); would be written as limit when x->0 of $$\frac{1}{y} sin(xy)$$; because f(0,y) equals $$\frac{1}{y} sin(xy)$$
as, limit when x->0 of $$\frac{1}{y} sin(xy)$$ equals 0
then also, limit when y->0 of 0 equals 0
so, result would be ZERO

Solution for (3):
isn't that only rewriting the two previous parts no more ?
and then to say the first limit equals the second one, so the final limit exists and equal to zero, right ?

13. Dec 31, 2008

### HallsofIvy

Staff Emeritus
Re: what is the mean of that in a "find limit" question !

Yes, that's true.

Yes, that is true.

No, you were told before- it is not enough that the limit given by two different paths are the same. You would have to prove that you get the same limit by all paths. And there is no easy way to do that. Either use the properties of limits that I mentioned before or convert to polar coordinates and take the limit as r goes to 0. Since only r measures the distance to (0,0), the limit exists as long as the limit as r goes to 0 does not depend on $\theta[/theta]? 14. Dec 31, 2008 ### AbuYusufEg Re: what is the mean of that in a "find limit" question ! i'm sorry but how to use the limits' properties ? i can't get you in your first reply about limits' properties ! 15. Dec 31, 2008 ### HallsofIvy Staff Emeritus Re: what is the mean of that in a "find limit" question ! The limit properties I mentioned (but, of course, I don't know what your course has covered so I don't know if you can use this) were $\lim_{(x,y)\rightarrow (a,b)} [f(x,y)+ g(x,y)]= \lim_{(x,y)\rightarrow (a,b)} f(x,y)+ \lim_{(x,y)\rightarrow (a,b)} g(x,y)$ and $\lim_{(x,y)\rightarrow(a,b)} f(x)g(y)= (\lim_{x\rightarrow a}f(x))(\lim_{y\rightarrow b} g(y)$ Using the first of those $$\lim_{(x,y)\rightarrow (0,0)} x^2+ y= \lim_{(x,y)\rightarrow (0,0)} x^2+ \lim_{(x,y)\rightarrow (0,0)} y$ By the second, [tex]\lim_{(x,y)\rightarrow (0,0)}x^2= \left(\lim_{x\rightarrow 0}x^2\right)\left(\lim_{y\rightarrow 0}1\right)= 0(1)= 0$$
and
[tex]\lim_{(x,y)\rightarrow (0,0)} y= \left(\lim_{x\rightarrow 0}1\right)\left(\lim_{y\rightarrow 0} y\right)= 1(0)= 0[/itex]
Put those together.

Last edited: Jan 1, 2009