# limit

• Mar 3rd 2008, 05:07 AM
heathrowjohnny
limit
If $\displaystyle f(x,y) = \begin{cases} x \sin \frac{1}{y} \ \ \text{if} \ y \neq 0 \\ 0 \ \ \ \ \ \ \ \ \ \text{if} \ y = 0 \end{cases}$, show that $\displaystyle \lim_{(x,y) \to (0,0)} f(x,y) = 0$ but that $\displaystyle \lim_{x \to 0} \left(\lim_{y \to 0} f(x,y) \right) \neq \lim_{y \to 0} \left(\lim_{x \to 0} f(x,y) \right).$

So for the first part, I simply plugged in $\displaystyle (x,y) = (0,0)$ and got $\displaystyle 0$. For the second part, I think the limits oscillate and don't exist?
• Mar 3rd 2008, 05:39 AM
CaptainBlack
Quote:

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

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

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

as:

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

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

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

does not exist

RonL