# Help with fucntion's Continuity check

• July 3rd 2010, 06:36 AM
5n1663r
Help with fucntion's Continuity check
Hi
How should I solve this:
Check these two functions continuity:

$
f(x,y)=
\left\{
\begin{array}{lr}
\frac {x^3-y^3}{x-y}&x \neq y\\
{x^2+xy+y^2}&x = y
\end{array}
\right.$

$
f(x,y)=
\left\{
\begin{array}{lr}
\frac {x^3y^2}{x^6+y^4}&(x,y) \neq (0,0)\\
0&(x,y) = (0,0)
\end{array}
\right.$

Thank you very much
• July 3rd 2010, 07:15 AM
CaptainBlack
Quote:

Originally Posted by 5n1663r
Hi
How should I solve this:
Check these two functions continuity:

$
f(x,y)=
\left\{
\begin{array}{lr}
\frac {x^3-y^3}{x-y}&x \neq y\\
{x^2+xy+y^2}&x = y
\end{array}
\right.$

This only needs checking at the point $x=y$ as it is continuous everywhere else. To check at this point you should use the factorisation:

$x^3-y^3=(x-y)(x^2+xy+y^2)$

CB
• July 3rd 2010, 07:20 AM
CaptainBlack
Quote:

Originally Posted by 5n1663r

$
f(x,y)=
\left\{
\begin{array}{lr}
\frac {x^3y^2}{x^6+y^4}&(x,y) \neq (0,0)\\
0&(x,y) = (0,0)
\end{array}
\right.$

If the limit as $x$ goes to zero from the right along the curve $y=x^{3/2}$ of $\frac {x^3y^2}{x^6+y^4}$ is different from zero then the function is not continuous.

CB
• July 3rd 2010, 08:03 AM
5n1663r
Thank you very much