# limit

• Mar 3rd 2008, 06:07 AM
heathrowjohnny
limit
If $f(x,y) = \begin{cases} x \sin \frac{1}{y} \ \ \text{if} \ y \neq 0 \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ y = 0 \end{cases}$, show that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$ but that $\lim_{x \to 0} \left(\lim_{y \to 0} f(x,y) \right) \neq \lim_{y \to 0} \left(\lim_{x \to 0} f(x,y) \right).$

So for the first part, I simply plugged in $(x,y) = (0,0)$ and got $0$. For the second part, I think the limits oscillate and don't exist?
• Mar 3rd 2008, 06:39 AM
CaptainBlack
Quote:

Originally Posted by heathrowjohnny
If $f(x,y) = \begin{cases} x \sin \frac{1}{y} \ \ \text{if} \ y \neq 0 \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ y = 0 \end{cases}$, show that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$ but that $\lim_{x \to 0} \left(\lim_{y \to 0} f(x,y) \right) \neq \lim_{y \to 0} \left(\lim_{x \to 0} f(x,y) \right).$

So for the first part, I simply plugged in $(x,y) = (0,0)$ and got $0$. For the second part, I think the limits oscillate and don't exist?

$\lim_{y \to 0} \left(\lim_{x \to 0} f(x,y) \right)=0$

as:

$\lim_{x \to 0} f(x,y)=0$

The problem is with the other limit as for any $x\ne 0$:

$\lim_{y \to 0} f(x,y)$

does not exist

RonL