Continuity of Functions with Unbounded Domains and the Limit of Compositions

[quasar987]

Is the following true? Is so, under what conditions, and what are, roughly, the arguments used to prove it?

f,g two functions of superiorly unbounded domain and such that for x > N, g(x) is continuous and f(g(x)) is continuous.

$$\lim_{x \rightarrow \infty} f(g(x)) = f(\lim_{x \rightarrow \infty} g(x))$$

I'm trying to show what

$$\lim_{x \rightarrow \infty} ln \left(\frac{x-1}{x+1} \right)=0$$

where ln is the natural logarithm (I think some people use the notation log for that). And without that "theorem", I don't see how to do it.

 

[matt grime]

log is continuous. (x-1)/(x+1) tends to 1 as x tends to infinity. you see how to put that together? (don't even know what superiorly bounded means...)

 

[quasar987]

superiorly unbounded: Is that not a word? I meant to say the domain has no upper bound.

I see how to put that together if the proposition I stated is true.

I guess you're saying it is, then.

 

[matt grime]

I have no desire to look at your proposition since:

a function is continuous at a point x if lim as y tends to x of f(y) is f(x). So your result is a consequnce of the fact that log is continuous at 1, that's all.

 

[quasar987]

So you're saying that if

$$\lim_{x \rightarrow \infty} f(x) = L$$

And if g(x) is continuous at L, then

$$\lim_{x \rightarrow \infty} g(f(x)) = \lim_{x \rightarrow L} g(x)$$

is that right?

I can see from the definition of continuity how it's true for x --> x0 an accumulation point of the domain of f and if f(x0) is an accumulation point of the domain of g but for x --> infinity, I wasn't sure.

 

[shmoe]

g doesn't have to be continuous at L for your last post to be true-it's just a statement of the limit of g, not how this limit relates to g(L).

You need continuity of g at L if you want to assert that

$$\lim_{x \rightarrow \infty} f(x) = L$$

implies

$$\lim_{x \rightarrow \infty} g(f(x)) = \lim_{x \rightarrow L} g(x)=g(L)$$

which is what you're using for this problem.

 

[quasar987]

Ok, so for my problem, since log is continuous at 1, all I have to do is prove that the function $f(x) = (x-1)/(x+1)$ as x approches infinity is 1. I have never done that. So if someone could check and see if these steps are correct, I'd be very grateful.

We want to show that $\forall \epsilon > 0$, there is an M such that for x element of the domain of f (i.e element of the real), x > M implies

$$\left |\frac{x-1}{x+1}-1 \right| < \epsilon$$

We see that

$$\left|\frac{x-1}{x+1}-1 \right| = \left|\frac{x-1}{x+1}-\frac{x+1}{x+1} \right| = \left|\frac{x-1-x-1}{x+1} \right| = \left|\frac{-2}{x+1} \right| = \left|\frac{2}{x+1} \right| = \frac{2}{\left|{x+1} \right|}$$

and

$$\frac{2}{\left|{x+1} \right|} < \epsilon \Leftrightarrow |x+1|>\frac{2}{\epsilon}$$

But $|x+1| \leq |x|+1$. Therefor, if we can show that

$$|x|+1>\frac{2}{\epsilon} \Leftrightarrow |x|>\frac{2}{\epsilon}-1 \ \forall \epsilon>0$$

we will have won. We can suppose without loss of generality that $M \geq 0 \Rightarrow x>0 \Rightarrow |x| = x$. So that choosing M = max{0, $\frac{2}{\epsilon}-1$} does the trick.

 

[shmoe]

You hit a snag when you applied the triangle inequality. Knowing $|x+1| \leq |x|+1$ will not help you get a lower bound for $|x+1|$. Try looking at $|x+1|\geq |x|-|1|$.

 

[quasar987]

Right! I did that part too mechanically. Thanks!