Use spherical coordinates to find the limit. [Hint: Let x = rho sin(ϕ) cos(θ), y = rho sin(ϕ) sin(θ), and z = rho cos(ϕ), and note that (x, y, z) → (0, 0, 0) implies rho → 0+.] lim (x, y, z) → (0, 0, 0) arctan 19 x2 + y2 + z2

lim (x, y, z) → (0, 0, 0) [x*y*z]/(x^2+y^2+z^2)=0

Step-by-step explanation:

First, we need to put the spherical coordinates equations that we will use in our problem  

x = p*sin∅cos Ф

y = p*sin∅sinФ

z=p*cos∅

Then we will state the problem  

lim (x, y, z) → (0, 0, 0)  [x*y*z]/(x^2+y^2+z^2)

Using the spherical coordinates we get  

(x^2+y^2+z^2) = p^2  

Which will make our limit be  

lim p→0+ [p*sin∅cos Ф*p*sin∅sinФ *p*cos∅]/(x^2+y^2+z^2)

after solving limit:

lim (x, y, z) → (0, 0, 0) [x*y*z]/(x^2+y^2+z^2)=0

146%

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.