Re: Limit

William Elliot schrieb:

On Thu, 22 Sep 2005, Antonia Senton wrote:

William Elliot schrieb:

On Wed, 21 Sep 2005, Antonia Senton wrote:

Jos? Carlos Santos schrieb:

anz3 wrote:


if g(a,f(a)) is a function continuous in both arguments. Is the
following equality valid:


\lim_{a \to 1} g(a,f(a)) = \lim_{a \to 1} g(a, \lim_{a \to 1} f(a)) ?

The limits of g and f exist. Why is that equality correct? Thanks! 

When you say that the limit of f exists, my guess is that what you
mean is that the limit lim_{a -> 1} f(a) exists. But exactly what do
you mean when you say that the limit of g exists?


Yes, I meant lim_{a -> 1} f(a) < \infty.


But exactly what do you mean when you say that the limit of g exists? 

The limit of the lhs exists, i. e. \lim_{a \to 1} g(a,f(a)) < \infty.

Moreover, I should mention that f is continuous in 1. 

Same counterexample.  Let g(x,y) = (x - 1)/y, f(x) = x - 1
lim(x->1) g(x,f(x)) = lim(x->1) (x - 1)/(x - 1) = 1
lim(x->1) f(x) = 0 = f(1);  f continuous at 1.
lim(x->1) g(x, lim(x->1) f(x)) = lim(x->1) g(x,0)
= lim(x->1) (x - 1)/0 = +-oo


That is not a counterexample since I stated above that g is assumed to
be continuous in both arguments (and g(x,y) = (x - 1)/y is not
continuous in g(x,1)). As I already stated the equality can probably be
proved by using that the composition of continuous functions is continuous.


You said that g(a,f(a)) is continuous in both arguments which
doesn't parse as g(a,f(a)) has only one argument.  If you mean
g(x,y) is continuous, then yes. g(x,f(x)) is continuous.

The following maps are continuous and compose to g(x,f(x))
x -> (x,x)
(x,y) -> (x,f(y))
g

In similar fashion, if h(x,y) and f(x), g(x)
are continuous, then h(f(x),g(x)) is continuous.

The function g itself has to arguments. If it is evaluated at (x,f(x)) it just could equivalently be written as a function of one variable. I think one has to consider x -> (x,f(x)) and (x, f(x)) -> g(x,f(x)). Then one knows that lim_{x->1} g(f(x)) = g(f(1)) (assuming that f is continuous in 1, and g in (1,f(1)) ).

.

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