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

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12