Let f (x, y) = x3 + y3 x2 + y2 . (a) Show that |x3|≤|x|(x2 + y2), |y3|≤|y|(x2 + y2) (b) Show that |f (x, y)|≤|x|+|y|. (c) Use the Squeeze Theorem to prove that lim (x, y)→(0,0) f (x, y) = 0.

Answer and Step-by-step explanation:

a.)x^2 and y^2 are always greater than zero.(assuming real x and y).

=>x^2<=x^2+y^2

=>|x|(x^2)<=|x|(x^2+y^2)

=>|x^3|<=|x|(x^2+y^2)

b. similarly, |y^3|<=|y|(x^2+y^2)

=>|f(x,y)|=|x^3+y^3|/(x^2+y^2)=|x|+|y|

c. let h is very small number (close to zero but positive)

lim (x,y)-> (0,0) f(x,y)=(x^3+y^3)/(x^2+y^2)

putting x=h and y=h and approaching h->0

lim h->0 f(h)=2h=0

There is another explanation attached below

37

step-by-step explanation:

idk

step-by-step explanation: