# Two-Variable Limit and Continuity

• Apr 6th 2010, 08:59 PM
ghow90
Two-Variable Limit and Continuity
I'm having trouble with this problem

f(x,y)= {(sin(x^2+y^2 )/((x^2+y^2 ) ) (x,y)=(0,0)
1 (x,y)≠(0,0)

The question says to show that f is continuous
• Apr 6th 2010, 09:10 PM
Prove It
Quote:

Originally Posted by ghow90
I'm having trouble with this problem

f(x,y)= {(sin(x^2+y^2 )/((x^2+y^2 ) ) (x,y)=(0,0)
1 (x,y)≠(0,0)

The question says to show that f is continuous

I assume that you actually mean

$f(x, y) = \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$ for $(x, y) \neq (0, 0)$

and $f(x, y) = 1$ for $(x, y) = (0, 0)$.

To show that this is continuous, you'd have to show that

$\lim_{(x, y) \to (0, 0)} f(x, y) = 1$.

$\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$

$= \lim_{r \to 0} \frac{\sin{r^2}}{r^2}$

since $r^2 = x^2 + y^2$.

It is well known that $\lim_{X \to 0} \frac{\sin{X}}{X} = 1$ (and can be shown using the sandwich theorem).

Therefore $\lim_{r \to 0} \frac{\sin{r^2}}{r^2} = 1$.

Since this limit is $1$, the function is continuous.
• Apr 6th 2010, 09:11 PM
ghow90
Quote:

Originally Posted by Prove It
I assume that you actually mean

$f(x, y) = \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$ for $(x, y) \neq (0, 0)$

and $f(x, y) = 1$ for $(x, y) = (0, 0)$.

To show that this is continuous, you'd have to show that

$\lim_{(x, y) \to (0, 0)} f(x, y) = 1$.

$\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}$

$= \lim_{r \to 0} \frac{\sin{r^2}}{r^2}$

since $r^2 = x^2 + y^2$.

It is well known that $\lim_{X \to 0} \frac{\sin{X}}{X} = 1$ (and can be shown using the sandwich theorem).

Therefore $\lim_{r \to 0} \frac{\sin{r^2}}{r^2} = 1$.

Since this limit is $1$, the function is continuous.

Thank You for the help
• Apr 7th 2010, 03:02 AM
HallsofIvy
Notice that Prove It changed to polar coordinates. That simplifies the problem greatly- since just r itself measures the distance to (0,0) regardless of $\theta$. As long as the limit, as r goes to 0, does NOT depend on [tex]\theta[\math], that gives the correct limit.