# Thread: Limit of multivariable function

1. ## Limit of multivariable function

I was just wondering if someone could give me short tutorial in limits of multi-variable functions... I understand that you approach the point from an arbitrary line y=cx. but i dont understand when my textbook, and many websites, explain limits using
0 < $\sqrt{(x-a)^2 + (y-b)^2}$ < δ

2. Recall for the limits of functions of one variable

|f(x)-L| < e whenever |x-a| < d or a - d < x < a + d

what you are saying is when the distance from any point x to a < d

then |f(x)-L| < e

It's the same basic idea for functions of 2 variables except instead of an interval centered at a you use a circle centered at (a,b)

so whenever the distance from (a,b) to (x,y) < d then |f(x,y)-L| < e

using the distance formula this takes the form

if 0 < < δ then |f(x,y)-L| < e

Be Careful approaching (a,b) along a line segment y =cx never proves a limit but if you appraoach along 2 different lines, or curves for that matter, and get different results then the limit does not exist.

hope this helps a little

3. yes thank you it does... so just one more question
when you have a function like:
$f(x,y) = ( \frac{x^2y^3}{2x^2+y^2}$if (x,y) DNE (0,0) )
( 1 if (x,y) = (0,0) )
and it asks where the function is continuous.

Would this just be everywhere except (0,0)???

4. What is helpful is to realize that sums, products, and quotients of continous functions of one one variable are continuous functions of 2 variables.
Example x^2 and y^3 are cont fns of one variable so x^2y^3 is a continuous function of two variables etc.

In your example as long a we stay away from (0,0) we have continuity.

Suppose you appraoach (0,0) along y = x then f(x,x) = x^5/(3x^2) whose limit is 0. This doesn't mean the limit at(0,0) is 0

The point is f(0,0) = 1 so even if the limit were 0 f would not be continuos.

In fact if you approach (0,0) along y= 1/x limf(x,y) doesn't even exist .

I hope this clarifies things a little.