Limit continuous function of rational numbers

[Krovski]

Homework Statement

Fix a real number a>1. If r=p/q is a rational number, we define a^r to be a^(p/q). Assume the fact that f(r)=a^r is a continuous increasing function on the domain Q of rational numbers r.

Let s be a real number. Prove that lim r--->s f(r) exists

Homework Equations

For all ε>o there exists δ>0 s.t. |x-x0|<δ implies |f(x)-f(x0)|<ε
Where x0 is an accumulation point for the domain

The Attempt at a Solution

Fix ε>0 and set δ=ε [not sure about this]
With s an accumulation point for Q
|r-s|<δ implies |f(r)-f(s)|<ε

[trouble here]
|f(r)-f(s)|= |a^r - a^s|
[i'm not sure how I should alter this equation to get to |r-s|<δ=ε. Perhaps I am taking the wrong approach? I just need to show that a limit exists as r--->s ]

 

[mighty2000]

Thank you for your post. I am a scientist and I can help you with your problem. Let's start by looking at the definition of a continuous function. A function f(x) is continuous at a point x0 if for any ε>0, there exists a δ>0 such that for all x in the domain, |x-x0|<δ implies |f(x)-f(x0)|<ε. In simpler terms, this means that as x gets closer and closer to x0, the values of f(x) get closer and closer to f(x0).

In your problem, s is an accumulation point for the domain of rational numbers, Q. This means that there exists a sequence of rational numbers, {r_n}, that converges to s. In other words, as n gets larger and larger, r_n gets closer and closer to s.

Now, we want to show that lim r--->s f(r) exists. This means that as r gets closer and closer to s, the values of f(r) get closer and closer to some number, which we will call L. In other words, for any ε>0, there exists a δ>0 such that for all r in the domain, |r-s|<δ implies |f(r)-L|<ε.

To prove this, we will use the definition of a continuous function. Fix ε>0 and let δ=ε. Now, we want to show that |f(r)-L|<ε for all r in the domain such that |r-s|<δ. We can rewrite this as |a^r-a^s|<ε. Now, we can use the fact that f(r)=a^r is a continuous, increasing function on the domain of rational numbers. This means that as r gets closer and closer to s, the values of f(r) will also get closer and closer to f(s). In other words, as r gets closer and closer to s, |f(r)-f(s)| will get smaller and smaller. This is exactly what we need to show, since |f(r)-f(s)| is equal to |a^r-a^s|. Therefore, we have shown that for any ε>0, there exists a δ>0 such that for all r in the domain, |r-s|<δ implies |f(r)-L|<ε. This means that lim r--->