# MA101.17 Continuity of functions of two variables

We can say that $f(x,y)\to L$ as the $(x,y) \to (a,b)$, or $\displaystyle \lim_{(x,y)\to (a,b)} f(x,y) = L$.

If, intuitively speaking, by going close enough to (a,b) we can get f(a,b) as close as we like to L.

f(x,y) is said to be continuous at (a,b) if $\displaystyle \lim_{(x,y)\to (a,b)} f(x,y) = f(a,b)$.

Just as with one variable limits from below and above may be different, with two variables we may get various limits coming in from different paths.

##### Example

$\displaystyle f(x,y)=\frac{xy}{x^2+y^2} \qquad (x,y) \neq (0,0)$

Consider coming in to (0,0) along the line y=mx (m fixed). Along this line we have:

$\displaystyle f(x,y) = f(x,mx) = \frac{xmx}{x^2 + m^2x^2} = \frac{m}{1+m^2}$

Thus along y=mx,

$\displaystyle \lim_{(x,y)\to (0,0)} f(x,y) = \lim_{x\to 0} \frac{m}{1+m^2} = \frac{m}{1+m^2}$

If we come in along the line $y=x^2$,

$\displaystyle \lim_{along y=x^2} f(x,y) = \lim_{x\to 0} \frac{x^3}{x^2+x^4} = \lim_{x\to 0}\frac{1}{\frac{1}{x}+x} = 0$.

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