Natural Logarithm and LimitsDate: 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) z->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 http://mathforum.org/dr.math/ |
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