# Multivariable calculus

Just wondering if anyone can prove these to me:

lim (x,y)->(a,b) f(x) * g(y) = lim x->a f(x) * lim y->b g(x) (As well as the n dimensional case)

Also, why when you try to show that a limit doesnt exist you can keep a variable constant, or do something like y=x, or approach from some other path and show that one path of approach doesnt give the same limit as another path of approach. This statement seems kind of ambiguous to me. Is there a rigorous way of proving this using the definition of the limit in Rn?

These are'nt specific questions of the textbook just things i am curious about and want to gain a better understanding

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HallsofIvy
Homework Helper
ak416 said:
Just wondering if anyone can prove these to me:
lim (x,y)->(a,b) f(x) * g(y) = lim x->a f(x) * lim y->b g(x) (As well as the n dimensional case)
Also, why when you try to show that a limit doesnt exist you can keep a variable constant, or do something like y=x, or approach from some other path and show that one path of approach doesnt give the same limit as another path of approach. This statement seems kind of ambiguous to me. Is there a rigorous way of proving this using the definition of the limit in Rn?
These are'nt specific questions of the textbook just things i am curious about and want to gain a better understanding
The definition of limit, $lim\x\rightarrow a f(x)= L$ in the ndimensional case is (f: Rn-> Rm, x and a are n-vectors, L is an m-vector):
Given $\epsilon> 0$ there exist a number $\delta> 0$ such that if ||x- a||< $\delta$ then $||f(x)- L||< \epsilon$. "|| ||" is, of course, some norm in Rn. One of the nice things about finite dimensional space, Rn, is that the norm, $||v||= max{|x_1|, |x_2|, ..., |x_n|}$ gives exactly the same "topology" (i.e. limits, etc.) properties as the "usual norm" $||v||= \sqrt{x_0^2+ x_1^2+ ...+ x_n^2}$. That means that you can look at the x and y components separately. That's why
$lim_{(x,y)\rightarrow (a,b)}f(x)g(y)= \left(lim_{x\rightarrowa}f(x)\right)\left(lim_{x\rightarrowb}g(y)$

The point of showing that a limit, as x-> a of f(x), does NOT exist by showing that limits taken in two different ways do NOT give the same result is due to the fact that saying the limit is, say, L, means that for any x sufficiently close to the a, f(x) is "close" to L.

If, by taking the limits in two different paths, you get two different limits, say L1 and L2, that means that there are points, x, y, as close as you wish to a, on each path, such that f(x) is close to L1 and f(y) is close to L2. Taking, say, $\epsilon= \frac{L_1-L2}{3}, shows that either x or y is does not give a value within [itex]\epsilon$.

thanks for the reply, i feel somewhat more confident now.

but does this mean that the limit exists if it is exists on at least one norm or it must exist for all norms? Im guessing if it exists on at least one norm it will exist on all norms..

HallsofIvy
Homework Helper
If your space is finite dimensional, you can show that the various norms:
$$||<x_1, x_2, ..., x_n>|| = \sqrt{x_1^2+ x_2^2+ ...+ x_n^2}$$
$$||<x_1, x_2, ..., x_n>|| = |x_1|+ |x_2|+ ... + |x_n|$$
$$||<x_1, x_2, ..., x_n>|| = max{|x_1|, |x_2|, ..., |x_n|}[/itex] Then, yes, a sequnce converges in all of those norms if and only if it converges in any one of them (and converges to the same thing in all norms.) However, there are norms corresponding to those in infinite dimensional spaces (such as function spaces) that give very different results. Even in finite dimensional spaces (even in R itself) one could use the "discrete" norm (||v||= 0 if v= 0, 1 otherwise) which gives the result that only constant sequences converge. benorin Homework Helper Gold Member Those norms are finite dimensional analogs of the norms in [tex]\ell ^{2},\ell ^{1},\mbox{ and }\ell ^{\infty},$$ correct? Do similar results apply to those spaces and, say $$\ell ^{p},L^{p},..$$? How about Hilbert or Banach spaces in general? I'm just learning about these spaces so that I am curious enough to ask (but should these have obvious answers, just tell me to go study harder.)

HallsofIvy
Homework Helper
benorin said:
Those norms are finite dimensional analogs of the norms in $$\ell ^{2},\ell ^{1},\mbox{ and }\ell ^{\infty},$$ correct? Do similar results apply to those spaces and, say $$\ell ^{p},L^{p},..$$? How about Hilbert or Banach spaces in general? I'm just learning about these spaces so that I am curious enough to ask (but should these have obvious answers, just tell me to go study harder.)
Yes those are the "infinite" dimensional spaces I was talking about. But notice that I said "However, there are norms corresponding to those in infinite dimensional spaces (such as function spaces) that give very different results."

In particular, L0(C) is defined as the set of all continuous function on compact set C with norm |f|= max|f(x)|, maximum taken over all x in C. That corresponds directly to the $|||| = max{|x_1|, |x_2|, ..., |x_n|}$ norm on Rn.

L1(C) is the set of (Lebesque) integrable functions on C with norm $$|f|= \int_C|f(x)|dx. That corresponds directly to the $|||| = |x_1|+ |x_2|+ ... + |x_n|$ norm. L2(C) is the set function whose square is (Lebesque) integrable on C with norm [tex]|f|= \sqrt{\int_C f^2(x)dx}$$. That corresponds directly to the "usual" norm$$|||| = \sqrt{x_1^2+ x_2^2+ ...+ x_n^2}$$

In general, Lp(c) is the set of functions whose pth power is (Lebesque) integrable on C with norm $$|f|= ^p\sqrt{\int_C f^p(x)dx}$$. There is the obvious analog on Rn but I've never seen it used except for p= 2.

However, while all three norms give exactly the same topology and so the same convergence on Rn, that is not true for their infinite dimensional analogs. The Lp spaces are quite different for different p.

As far as Hilbert and Banach spaces in general are concerned, the inner product and norm are part of the definition. Exactly the same set with different norms defined will, in general, be different Banach spaces.

benorin
Homework Helper
Gold Member
Thanx Ivy.