Answer to Question #105360 in Calculus for Khushi

Question

State whether the following statements are true or false. Give reasons for your answers
(1) The function f:R^3®R, given by f( x, y, z) =|x|+|y|+|z| is differential at (2, 3,-1).
(2) The function f(x, y) =max{y/x, x} is a homogenous function on R^2.
(3) The domain of the f/g where f(x, y) =2xy and g(x, y) =x^2+y^2 is R^2.

Answer 1

1)It is true. In $B_1(2,3,-1)=\{(x,y,z)\mid |(2,3,-1)-(x,y,z)|<1\}$ we have $f(x,y,z)=x+y-z$. It is a differentiable in $B_1(2,3,-1)$ function, so it is differentiable in $(2,3,-1)$.

2)It is not true. For $x_0=1,y_0=1,\lambda_0=0$ we have $\lambda_0^kf(x_0,y_0)=0^k\max\{\frac{1}{1},1\}=1$ is defined, but $f(\lambda_0x_0,\lambda_0y_0)=\max\{\frac{\lambda_0y_0}{\lambda_0x_0},\lambda_0x_0\}=\max\{\frac{0}{0},0\}$ is not defined, because $\frac{0}{0}$ is not defined, so $\lambda_0^kf(x_0,y_0)\neq f(\lambda_0x_0,\lambda_0y_0)$.

3)It is false, beacuse $\frac{f}{g}$ is not continuous at $(0,0)$, so we cannot define $\frac{f}{g}$ at this point.

$\frac{f}{g}$ is not continuous at $(0,0)$, because there is not a $\lim\limits_{(x,y)\to (0,0)} \frac{f(x,y)}{g(x,y)}$. Indeed, for $\varepsilon=1$ we have that for every $\delta>0$ we can take $\left(0,\frac{\delta}{2}\right),\left(\frac{\delta}{2},\frac{\delta}{2}\right)\in B_{\delta}(0)=\{(x,y)\mid |(x,y)|<\delta\}$ and $\left|\frac{f\left(0,\frac{\delta}{2}\right)}{g\left(0,\frac{\delta}{2}\right)}-\frac{f\left(\frac{\delta}{2},\frac{\delta}{2}\right)}{g\left(\frac{\delta}{2},\frac{\delta}{2}\right)}\right|=|0-1|\ge\varepsilon$, so $\frac{f}{g}$ is not satisfy the Cauchy test of limit existence at $(0,0)$.