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Math Help - Help with fucntion's Continuity check

  1. #1
    Newbie
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    Question Help with fucntion's Continuity check

    Hi
    How should I solve this:
    Check these two functions continuity:

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

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

    Thank you very much
    Last edited by 5n1663r; July 3rd 2010 at 06:54 AM.
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  2. #2
    Grand Panjandrum
    Joined
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    4
    Quote Originally Posted by 5n1663r View Post
    Hi
    How should I solve this:
    Check these two functions continuity:

    <br />
f(x,y)= <br />
\left\{<br />
\begin{array}{lr}<br />
\frac {x^3-y^3}{x-y}&x \neq y\\<br />
{x^2+xy+y^2}&x = y<br />
\end{array}<br />
\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
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  3. #3
    Grand Panjandrum
    Joined
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    someplace
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    4
    Quote Originally Posted by 5n1663r View Post

    <br />
f(x,y)= <br />
\left\{<br />
\begin{array}{lr}<br />
\frac {x^3y^2}{x^6+y^4}&(x,y) \neq (0,0)\\<br />
0&(x,y) = (0,0)<br />
\end{array}<br />
\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
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  4. #4
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    Thank you very much
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