# Thread: Show discontinuity in f(x,y)

1. ## Show discontinuity in f(x,y)

I have f(x,y) = 2xy^2 / (x^2+y^4) if (x,y) != (0,0) and 0 if (x,y) = (0,0).
I figured I could just find that the limit as 2xy^2 / (x^2+y^4) does not equal to 0, which should show discontinuity, but it does and I don't know how else one can show discontinuity. I showed it was not differentiable with the definition of the derivative if that helps at all, but I can't reuse that. Thanks for any help.

2. ## Re: Show discontinuity in f(x,y)

For f to be continuous at (0,0), three things must hold:

1. The point $\displaystyle (0,0)$ is in the domain of $\displaystyle f$. (It is: $\displaystyle f(0,0) = 0$.)

2. $\displaystyle \lim_{P \rightarrow (0,0)} f(P)$ exists.

3. $\displaystyle \lim_{P \rightarrow (0,0)} f(P) = f(0,0) = 0$.

Now, look at $\displaystyle P_n = \left(\frac{1}{n^2}, \frac{1}{n} \right)$.

Clearly have that $\displaystyle \lim_{n \rightarrow \infty} P_n = (0,0)$.

What is $\displaystyle \lim_{n \rightarrow \infty} f(P_n) ?$

3. ## Re: Show discontinuity in f(x,y)

Hey Shanter.

In this problem, you should look at showing what the limits are of the analytic portion from the right and from the left. Try doing two limits: one where x approaches zero and y is held constant and one where y approaches zero and x is held constant.

In the above case you will have to use L'Hopitals rule you should get the limit where y varies to go to infinity for any non-zero fixed x.

The definition of continuity is when the limit of a function for it's input approaching some point is equal to the value of the function at that point: in other words lim (x->a,y->b) f(x,y) = f(a,b) and if you can show that any limit does not equal the value of the function from any direction towards the origin, then you have shown that it is discontinuous.