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Since I usually studied and checked differentiability and continuity of one variable, and only a few two variable functions, proceeding to the chapter of multivariables, I encountered a problem.

One of the problems for practice says: Check the differentiability (alongside continuity) of a piecewise function: $$f(x,y,z) = \begin{cases} (x^2+y^2+z^2)\sin(\frac{1}{x^2+y^2+z^2}), & (x,y,z)\neq(0,0,0) \\[2ex] 0, & (x,y,z)=(0,0,0) \end{cases}$$ We usually always checked continuity aswell even though it might suffice to prove differentiability only.

Continuity in one var. functions was proved by checking limits or by the epsilon-delta definition. We proved differentiability by checking if both left and right derivatives (by definition) are equal.

I'm a bit confused here on both continuity and differentiability, which method to use. I don't know how I'd go about continuity. And for differentiability I suppose if all partial derivatives exist, the function is differentiable. So I'd have to check the equality of all sides of partial derivatives with limits? How is this usually handled?

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  • $\begingroup$ Note that existence of partial derivatives does not imply differentiability, specifically, partial derivatives need to be continuous. Do you have a limit/linear transformation definition of differentiability in multiple variables? $\endgroup$ – rubikscube09 May 5 '18 at 6:22
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The function $f$ is differentiable at $(0,0,0)$ and $f'(0,0,0)$ is the null function. What this means is that$$\lim_{(x,y,z)\to(0,0,0)}\frac{(x^2+y^2+z^2)\sin\left(\frac1{x^2+y^2+z^2}\right)}{\|(x,y,z)\|}=0,$$which is true, since the previous limit is equal to$$\lim_{(x,y,z)\to(0,0,0)}\|(x,y,z)\|\sin\left(\frac1{x^2+y^2+z^2}\right)$$and the sine function is bounded.

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