# Thread: Continuity of a 2 variables function

1. ## Continuity of a 2 variables function

Hello,

I need a hand with an exercise, it asks me to study the continuity of the following function for each value of "a" (sorry I don't know how to use LaTex!):

f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if y != 0
f(x,y) = a, if y = 0

Basically I don't how to solve the limit for y -> 0

Any idea?

Thanks in advance.

2. Originally Posted by jollysa87
Hello,

I need a hand with an exercise, it asks me to study the continuity of the following function for each value of "a" (sorry I don't know how to use LaTex!):

f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if y != 0
f(x,y) = a, if y = 0

Basically I don't how to solve the limit for y -> 0

Any idea?

Thanks in advance.
If what you're asking is

$
\lim_{(x,y) \to (0,0)} \frac{x^2 e^x}{y \,e^{y^2}}
$

try letting $x = r \cos \theta$ and $y = r \sin \theta$ and then let $r \to 0$ .

3. I already did it but the problem is that x can be anything, otherwise the exercise would have been:

f(x,y) = (x^2 * e^x) / [y * e^(y^2)], if (x,y) != 0
f(x,y) = a, if (x,y) = 0

Am I wrong??

4. So the problem is not $\lim_{(x,y)\rightarrow (0,0)} \frac{x^2e^x}{ye^{y^2}}$, it is $\lim_{y\rightarrow 0}\frac{x^2e^x}{ye^{y^2}}$?

Well, that's easier! If x is not 0, the denominator goes to 0 while the numerator does not so there is no limit. If x= 0, the function is 0 for all y and so the limit is 0.

5. so i can say that the limit for x different from 0 doesn't exist simply because the denominator goes to 0 while the numerator does not or I have to make some better explanation? thanks!

6. Originally Posted by jollysa87
so i can say that the limit for x different from 0 doesn't exist simply because the denominator goes to 0 while the numerator does not or I have to make some better explanation? thanks!
Yes, that is sufficient. The case where both x and y approach zero is really the only interesting case.