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Here is an example with your function. Fix $\alpha=1$ and imagine in rectangular coordinates, approaching the limit along the $x$-axis (so $y=0$). Then you get $$ \frac{x^2}{x^2+y^2} \to \frac{x^2}{x^2+0} = 1. $$ Now imagine approaching from the $y$-axis, so $x=0$. You get $$ \frac{x^2}{x^2+y^2} \to \frac{0}{0+y^2} = 0. $$ Since the values disagree, the ...


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There's no upper bound for the average speed. The speed is determined by the height; the deeper you go, the higher you can make the average speed.


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Consider the two points at which the tangent planes touch the paraboloid: $\left(x_1, y_1,\frac{x_1^2+y_1^2}{2}\right)$ and $\left(x_2, y_2,\frac{x_2^2+y_2^2}{2}\right)$ The tangent planes to these points are: $z-\frac{x_1^2+y_1^2}{2}=x_1(x-x_1)+y_1(y-y_1)$ and $z-\frac{x_2^2+y_2^2}{2}=x_2(x-x_2)+y_2(y-y_2)$. (This is from the equation $z-z_0=f_x(x_0,...


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Let the tangent planes touch the surface at $(x', y',z')$ and $(x'',y'',z'')$, $$\left \{ \begin{array}{rcl} x'x+y'y &=& z+z' \\ x''x+y''y &=& z+z'' \end{array} \right.$$ The line of intersection $(\xi, \eta, \zeta)$ is $$(\xi, \eta, \zeta)= \left( \frac{\begin{vmatrix} \zeta+z' & y' \\ \zeta+z'' & y'' \end{vmatrix}} ...


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The answer to your question hinges on what exactly $\alpha$ is supposed to be. If $\alpha$ is a parameter which defines a family of functions, there may be some values of $\alpha$ for which the function is continuous everywhere. It certainly is not continuous for all $\alpha$ in either $\mathbb{R}$ or $\mathbb{Z},$ since $\alpha = 0$ produces the function $$ ...



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