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 th
e Squeeze Theorem to prove that lim (x,y)→(0,0) f (x, y) = 0.
1 answer:
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
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F you just want to see the really short way, just skip down to AAAAAAAAAAAA
so, here is the long explanation
exponential properties
don't forget pemdas
2x^2=2(x^2)
so
=
AAAAAAAAAAAAAAAAAAAAAAAA
so we see
the original equatio is
remember
so we can seperate the constants
we know that the placeholders cannot affet the position of the constants unless they are grouped together which they are not
terfor the answer must have 2/3 in it
the only one that hsa that is A
ANSWER IS A
so, here is the long explanation
exponential properties
don't forget pemdas
2x^2=2(x^2)
so
AAAAAAAAAAAAAAAAAAAAAAAA
so we see
the original equatio is
remember
so we can seperate the constants
we know that the placeholders cannot affet the position of the constants unless they are grouped together which they are not
terfor the answer must have 2/3 in it
the only one that hsa that is A
ANSWER IS A