# Help with fucntion's Continuity check

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

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

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

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

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

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

This only needs checking at the point $\displaystyle $$x=y as it is continuous everywhere else. To check at this point you should use the factorisation: \displaystyle x^3-y^3=(x-y)(x^2+xy+y^2) CB • Jul 3rd 2010, 07:20 AM CaptainBlack Quote: Originally Posted by 5n1663r \displaystyle f(x,y)= \left\{ \begin{array}{lr} \frac {x^3y^2}{x^6+y^4}&(x,y) \neq (0,0)\\ 0&(x,y) = (0,0) \end{array} \right. If the limit as \displaystyle$$ x$ goes to zero from the right along the curve $\displaystyle y=x^{3/2}$ of $\displaystyle \frac {x^3y^2}{x^6+y^4}$ is different from zero then the function is not continuous.

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