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Given the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined $x,y \in \Bbb R$ and $x^2 \neq y^2$. Let $a \neq 0$, determine wether the $\lim \limits_{(x,y) \rightarrow (a,a)} f(x,y)$ exists or not. If it does exist give the value.

I thought you had to do this by looking at different paths, so I looked at $(x,y)=(t,bt)$ with $t \rightarrow a$. $\lim \limits_{t \rightarrow a} f(t,bt)$=$\lim \limits_{t \rightarrow a} \frac{t^3-b^3t^3}{t^2-b^2t^2}$=$\frac{b^2a+ba+a}{b+1}$.

I also did this for $(x,y)=(t,t^2)$ with $t \rightarrow a$. $\lim \limits_{t \rightarrow a} f(t,t^2)$= $\lim \limits_{t \rightarrow a} \frac{t^3-t^6}{t^2-t^4}$=$ \frac {a^3+a^2+a}{a+1}$.

Since $\lim \limits_{t \rightarrow a} f(t,bt) \neq \lim \limits_{t \rightarrow a} f(t,t^2)$ it doesn't exist.

I'm not sure if this is right, or how I could do this otherwise.

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    You can't do that,because both functions must approach the same value i.e if you approach by $f(t,bt)$ then $b=1$ or $b\to 1$.The $f(t,t^2)$ is not allowed except if $a=1,0$ – 2017-02-17
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    Plugging $b=1$ in your first limit gives you a solution to your limit. – 2017-02-17

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HINT: For $x\not=y$ $$\frac{x^3-y^3}{x^2-y^2}=\frac{x^2+xy+y^2}{x+y}$$ Now taking $\lim_{(x,y)\to(a,a)}$ should be easy because the function is continuous.

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    When i have to prove this exists I can't use that the function is continuous right? Because to get to that function you used $(x-y) \neq 0$. – 2017-02-18
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    @Zoey It's a valid step to cancel $x-y$ since $x\neq y$ when taking $\lim$ and when you cancel it the resulting function is continuous (except for x=-y), it doesn't matter if the original function was discontinuous – 2017-02-18
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Yes the limit exists and is equal to $1.5a^2$