In my textbook it's give that $|r(\cos^3\theta - \sin^3\theta)| < 2r$. I didn't get how 2 came as maximum value of $\cos$ and $\sin$ is $1$ so that $\cos^3\theta - \sin^3\theta$ should be 0 ( i.e, 1-1). Also for differentiability i used $df = Ah+Bk+h\phi+k\psi$. My professor told me to take $h = p\cos\theta$ and $k = p\sin\theta$ and then use for arbitrary $\theta = \tan^{-1}\frac{h}{k}$. that implies p tends to 0 as (h,k) tends to 0. I don't know how to use h and k and how to proceed.
Show that $\frac{x^3-y^3}{x^2+y^2}$ is continuous but not differentiable at origin.
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multivariable-calculus
derivatives
continuity
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0$|r(\cos^3\theta - \sin^3\theta)| < |r|(|\cos^3\theta|+| \sin^3\theta|)
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1There is no $\theta$ such that $|\cos^3\theta-\sin^3\theta|=2$ but $|\cos^3\theta-\sin^3\theta|\leqslant2$ for every $\theta$, is an easy upper bound (easy, since $|\cos\theta|\leqslant1$ and $|\sin\theta|\leqslant1$), probably sufficient to your textbook's purpose. (The second half of your post is difficult to decipher.) – 2017-02-01
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0Thanks i got the 1st half and For the second half if i have to prove that the given function is not differentiable then what my approach should be? – 2017-02-01
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0http://math.stackexchange.com/questions/2123956/show-that-fracx3-y3x2y2-is-not-differentiable-at-origin – 2017-02-01