We consider the 6-dimensional complex projective space of non-zero binary sextic forms
in x and y. Then the group $SL_2(C)$ acts here by change of coordinates. Hence any binary form has a corresponding SL_2-orbit. The SL_2-orbit of $x^4y^2$ is 2 dimensional, and the SL_2-orbit of x^6+y^6 is 3 dimensional.
Dear Alexandru, i asked Bruce Reznick and Alessandro Oneto. Hopefully they know the answer
I believe that the answer is "no". The form x^6+y^6 has a unique quadratic form that is apolar to it, i.e. whose differential sends x^6+y^6 to 0. This quadratic form is xy, as differentiating with respect to x then y sends x^6+y^6 to 0.
As we act by SL_2(C) we get sums of 2 powers of linear forms, each such form will have a unique (up to multiplication by constant) quadratic form apolar to it. This quadatic form will have 2 distinct roots, and the two roots give the decomposition as a sum of 2 powers of linear forms by apolarity lemma.
When we take limits, I believe the limit form still must have an apolar quadratic form, the difference is that the apolar quadratic may now have a double root. However, the apolar ideal of x^4y^2 is generated by x^5 and y^3 and does not contain a quadratic.
I hope this helps (and is also correct!).
Best,
Greg
A simple Gröbner basis computation shows that the answer is negative:
https://gist.github.com/mohamed-barakat/5c9db11bed1d4b0d9932cc3b36f9e08c
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