Help with fucntion's Continuity check

**5n1663r** · July 3rd 2010, 06:36 AM

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

**CaptainBlack** · July 3rd 2010, 07:15 AM

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

**CaptainBlack** · July 3rd 2010, 07:20 AM

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

**5n1663r** · July 3rd 2010, 08:03 AM

Thank you very much

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