rigin is the line $y=2x$. If we stay on
    this line, then we can replace each $y$ with $2x$ and then
    compute $$\ds\lim_{\substack{(x,y)\to(0,0)\\ y=2x
    }}\frac{x^2-y^2}{x^2+y^2} = \lim_{x\to 0}
    \frac{x^2-(2x)^2}{x^2+(2x)^2} = \lim_{x\to 0}
    \frac{-3x^2}{5x^2} = \lim_{x\to 0} \frac{-3}{5}
    =\frac{-3}{5}.$$ This means that if we approach the origin
    along the line $y=2x$, we will have a height of $-3/5$ when we
    arrive at the origin.
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    <p>If the function $\ds f(x,y)=\frac{x