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Natural Logarithm and Limits

Date: 03/25/99 at 02:28:19
From: Lindsay Johnson
Subject: Natural Logarithm and Limits

We just learned in my Math31IB class that the limit of a ln equals the  
ln of the limit. [i.e. lim (ln x) = ln (lim x)]. However, our teacher 
has tried unsuccessfully to find a proof for this, and we would love 
to see some proof that this is true.

Date: 03/25/99 at 11:55:51
From: Doctor Rob
Subject: Re: Natural Logarithm and Limits

The proof of this comes in two parts. The first part is to prove that
ln(x) is a continuous function. The second part is to prove that if f
is any continuous function at L = lim g(x), then

   lim f(g(x)) = f(lim g(x)),
   x->a            x->a

provided both limits exist.

The first part is pretty easy, and you probably already know how to do
that, so I will skip it.

For the second part we need the definition of a limit and the 
definition of a continuous function:

Def: The limit of F(z) as z approaches c equals L means that for every 
epsilon > 0, there exists a delta > 0 such that |z-c| < delta implies 
|F(z)-L| < epsilon.

Def: A function F(z) is continuous at a point z = c mean that F(c) is
defined, and that the limit as z approaches c of F(z) exists, and that
they are equal:

   lim F(z) = F(c)

In other words, for every epsilon > 0, there exists a delta > 0 such 
that |z-c| < delta implies |F(z)-F(c)| < epsilon.

Now to prove that

   lim f(g(x)) = f(lim g(x))
   x->a            x->a

start with the fact that L = lim g(x) exists, by hypothesis, and 
f is continuous at L, so the right side is defined. The left-side 
limit also exists, by hypothesis. Then you have to show that the two 
sides are equal. In other words, you have to show that for every 
epsilon > 0, there exists a delta > 0 such that |x-a| < delta implies 
|f(g(x))-f(L)| < epsilon. 

Let y = g(x). Then since f is continous at L, given epsilon > 0 
we can find a delta1 > 0 such that |y-L| < delta1 implies 
|f(y)-f(L)| < epsilon. Now since lim g(x) exists, we can find a 
delta > 0 such that |x-a| < delta implies |g(x)-L| < delta1, so 
|y-L| < delta1, so |f(g(x))-f(L)| < epsilon.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Analysis

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