Yasmin

2021-10-31

Determine if f is continuous at (0,0) where:
$f\left(x,y\right)=\left\{\begin{array}{ll}\frac{3{x}^{2}y}{{x}^{4}+2{y}^{2}},& \left(x,y\right)\ne \left(0,0\right)\\ 0& \left(x,y\right)=\left(0,0\right)\end{array}$

### Answer & Explanation

Anonym

If a function f is continuous, then $\underset{\left(x,y\right)\to \left(a,b\right)}{lim}f\left(x,y\right)$ is unique.
Here, in this problem we will show that $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}f\left(x,y\right)$ is not unique.
And hence, if is not continous.
Solution:
$f\left(x,y\right)=\left\{\begin{array}{ll}\frac{3{x}^{2}y}{{x}^{4}+2{y}^{2}},& \left(x,y\right)\ne \left(0,0\right)\\ 0& \left(x,y\right)=\left(0,0\right)\end{array}$
Let $y\to 0$ along the path $y=m{x}^{2}$ where
Now $\underset{x\to 0}{lim}f\left(x,y\right)\phantom{\rule{1em}{0ex}}y=m{x}^{2}$
$=\underset{x\to 0}{lim}\frac{3{x}^{2}y}{{x}^{4}+2{y}^{2}}$
$=\underset{x\to 0}{lim}\frac{3{x}^{2}\left(m{x}^{2}\right)}{{x}^{4}+2{\left(m{x}^{2}\right)}^{2}}$
$=\underset{x\to 0}{lim}\frac{3m{x}^{4}}{{x}^{4}+2{m}^{2}{x}^{4}}$
$=\underset{x\to 0}{lim}\frac{3m\frac{{x}^{4}}{{x}^{4}}}{\frac{{x}^{4}}{{x}^{4}}+2{m}^{2}\frac{{x}^{4}}{{x}^{4}}}$
$=\underset{x\to 0}{lim}\frac{3m}{1+2{m}^{2}}$
$=\frac{3m}{1+2{m}^{2}}$
Hence, for different values of m, the limit is different.
Therefore
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}f\left(x,y\right)$ is not unique.
Hence, $f\left(x,y\right)$ is not continuous.

Jeffrey Jordon