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Consider the family of functions $\{f_c \mid c\in\bf{R}\}$ where $$ \begin{align*} f_c &\colon{\bf R}^2\to{\bf R}\\ &f_c(x,y)=1-\big(x^2+4y^2\big)^c. \end{align*} $$ I intuitively see that this function is not continuous for $c\leq 0$, but how can I prove this with a 'clean' argument involving limits? How can I determine for which $c$ the function $f_c$ is differentiable?

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@Mark Dominus You are both right, I've just fixed it –  Mart G. Mar 26 '12 at 17:45
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Hint: first look at what happens as $x \to 0$ with $y=0$. –  Robert Israel Mar 26 '12 at 18:16
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For $c\geq 0$, $f_c$ is continuous at each point. For $c<0$, $f_c$ has no limit at $(0,0)$ since $\lim_{x\to 0}x^{2c}=+\infty$. Now we look at differentiability (for $c\geq 0$). Since $f_c\equiv 1$ we only look at $c>0$. We have $\frac{f_c(h,0)-f(0,0)}h=\frac{- h^{2c}}h=-h^{2c-1}$ which converges to $0$ as $h\to 0$ if $c>1/2$, equal to $-1$ if $c=1/2$ and there is no limit if $c<1/2$. Similarly, $f_c$ has a partial derivative with respect to $y$ if $c\geq 1/2$, which is equal to $0$ if $c>1/2$ and to $-2$ if $c=1/2$. For $c>1/2$ $$\left|\frac{f_c(x,y)-f_c(0,0)-x\partial_xf_x(0,0)-y\partial_y(0,0)}{\sqrt{x^2+y^2}} \right| =\frac{(x^2+4y^2)^c}{\sqrt{x^2+y^2}}\leq 4^c(x^2+y^2)^{c-1/2}$$ which converge to $0$ as $(x,y)\to (0,0)$. For $c=1/2$, $$\frac{f_c(x,y)-f_c(0,0)-x\partial_xf_x(0,0)-y\partial_y(0,0)}{\sqrt{x^2+y^2}}= \frac{-\sqrt{x^2+4y^2}+x+2y}{\sqrt{x^2+y^2}}$$ and look whether the limit of the RHS as $(x,y)\to (0,0)$ is $0$.

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My answer is a bit pedantic, but, as an instructor, I believe that these things should not happen. If $c \geq 0$, the function $f_c$ is continuous and differentiable, since it is composed of elementary differentiable functions.

If $c<0$, the function $f_c$ is undefined at $(0,0)$. It is therefore meaningless to check whether $f_c$ is continuous (or differentiable) at points that do not belong to the domain of definition. We may wonder whether $f_c$ can be extended to a continuous (and differentiable) function, and Davide Giraudo solved this problem in his answer.

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