What is the Epsilon-Delta Proof for the Dot Product of a Vector?

Question

Well, for my multivariate calculus math homework, I have to do the proof of this:
lim x->a (u(x)∙v(x)) = lim x->a (u(x)) ∙ lim x->a (v(x))

I have to use the epsilon-delta way to prove it and have no clue where to start.

Answer 1

Use the Schwarz Lone Starr. Use the Schwarz.

The Cauchy-schwarz inequality that is -- it states that the magnitude of the inner product of two vectors is less than or equal to the product of their magnitudes. With that inequality in hand, the rest of this is just analogous to the proof that ordinary multiplication is continuous. The proof follows:

Let U = [x→a]lim u(x) and V = [x→a]lim v(x). Then we wish to show that:

[x→a]lim () =

Let ε>0. From the definition of limit, ∃δ₁>0 such that 0<|x-a|<δ₁ ⇒ ||u(x) - U|| < min (ε/(||U|| + ||V|| + 1), 1) and ∃δ₂>0 such that 0<|x-a|<δ₂ ⇒ ||v(x) - V|| < min (ε/(||U|| + ||V|| + 1), 1). So now let δ = min (δ₁, δ₂). Then suppose that 0<|x-a|<δ. Then we will have:

| - |

= | - |

Using the linearity of the inner product:

= | + - |

= | + + + - |

= | + + |

Using the triangle inequality:

≤ || + || + ||

Using the Cauchy - Schwarz inequality:

≤ ||u(x) - U|| * ||v(x) - V|| + ||u(x) - U|| * ||V|| + ||U|| * ||v(x) - V||

Now, since 0<|x-a|<δ≤δ₁, we have that ||u(x) - U|| < min (ε/(||U|| + ||V|| + 1), 1), so in particular ||u(x) - U|| < ε/(||U|| + ||V|| + 1). Also, since 0<|x-a|<δ≤δ₂, ||v(x) - V|| < min (ε/(||U|| + ||V|| + 1), 1), so in particular ||v(x) - V|| < ε/(||U|| + ||V|| + 1) and also ||v(x) - V|| < 1. So using these inequalities on the above:

< ε/(||U|| + ||V|| + 1) * 1 + ε/(||U|| + ||V|| + 1) * ||V|| + ε/(||U|| + ||V|| + 1) * ||U||

And now just factoring:

= ε/(||U|| + ||V|| + 1) * (1 + ||V|| + ||U||)

= ε

Therefore, we have that 0<|x-a|<δ ⇒ | - | < ε. Since we can find such a δ for any ε>0, it follows that [x→a]lim = . Q.E.D.

Answer 2

The (perspective A - perspective B) is the perspective between A and B. in view that vectors are geometrical products, we are able to pass them around as we see greater healthful. enable's place the vectors A and B such that they stem from the middle of a circle; i.e. tails initiate at center and element outward from the middle. to discover the perspective between A and B, we subtract the angles each and every makes in connection with the cirlce. as an occasion: enable's say that A makes an perspective of fifty ranges and B makes an perspective of 20 ranges. the perspective between A and B is: 50 - 20 = 30 ranges. desire this facilitates.