Calculus help

[junhoma]

Dear sir,Madam
Can you please help me to solve this function.

1. Compare and contrast the continuity of the functions f, g, h defined below.

f(x, y) = 2x^2y^2/(x^2+y^2)
g (x,y)=2x^2y^2/(x^2+y^2) if (x,y) is not equal to (0,0) and 0 if (x,y)=(0,0)
h(x,y)= 2x^2y^2/(x^2+y^2) if (x,y) is not equal to (0,0) and 2 if (x,y)=(0,0)
please sir help me.

 

[galactus]

Continuity can be extended to two variables as it is with one variable.

We know f(x) is continuous if \(\displaystyle \lim_{x\to x_{0}}f(x)=f(x_{0})\)

Therefore:

\(\displaystyle \lim_{(x,y)\to (x_{0},y_{0})}f(x,y)=f(x_{0},y_{0})\)

\(\displaystyle f(x, y) = \frac{2x^{2}y^{2}}{x^{2}+y^{2}}\) is continuous everywhere, except where \(\displaystyle x^{2}+y^{2}=0\)

So, \(\displaystyle \lim_{(x,y)\to (0,0)}\frac{2x^{2}y^{2}}{x^{2}+y^{2}}=0\)

You can try the limits from different paths, as in the previous posts, and see that it equals 0 each time.

Did you know that polar methods may also be used to show \(\displaystyle \lim_{(x,y)\to (0,0)}\frac{2x^{2}y^{2}}{x^{2}+y^{2}}=0\) by letting \(\displaystyle x^{2}+y^{2}=r^{2}\)?.

If this is the case, then \(\displaystyle x^{2}\leq r^{2} \;\ \text{and} \;\ y^{2}\leq r^{2}\)

\(\displaystyle \left|\frac{2x^{2}y^{2}}{x^{2}+y^{2}}-0\right|=\frac{2x^{2}y^{2}}{x^{2}+y^{2}}\leq \frac{2r^{4}}{r^{2}}=2r^{2}< {\epsilon} \;\ \text{if} \;\ r<\sqrt{\frac{\epsilon}{2}}\). So, choose \(\displaystyle {\delta}=\sqrt{\frac{\epsilon}{2}}\).

This is probably more rigor than you need, but it can be used to prove the limit of f(x) = 0.