What's the general method to find the slope of a curve at the origin if the derivative at the origin becomes indeterminate. For Eg--
What is the slope of the curve $x^3 + y^3= 3axy$ at origin and how to find it because after following the process of implicit differentiation and plugging in $x=0$ and $y=0$ in the derivative we get $0/0$.
Actually this question has been asked by me before and a sort of satisfactory answer that I got was
" For small $x$ and $y$, the values of $x^3$ and $y^3$ will be much smaller than $3axy$, so the zeroes of the function will be approximately where the zeroes of $0=3axy$ are -- that is, near the origin the curve will look like the solutions to that, which is just the two coordinate axes. So the curve will cross itself at the origin, passing through the origin once horizontally and once vertically. (This is also why implicit differentiation can't work at the origin -- the solution set simply doesn't look like a straight line there under any magnification)."
If I approximate the function by saying that at (0,0) , the behavior is dominated be 3axy term as x^3 and y^3 are very small and then 3axy=0 and then tangents are x=0 and y=0 . Is doing so (saying x=0 and y=0) linear Approximation only. Because I am approximating the curve with a straight line at origin . But linear Approximation is 1st derivative (1st term of Taylor series). This cannot be right because Taylor series can't be formed where derivative doesn't exist*.
And if this is right then the function is approximately given by 3axy=0 at (0,0). But how does this give the tangent at (0,0).How shall I go about ?
Is the answer give right because the solpe does exist.