# Is this proposition true?

1. Dec 2, 2004

### 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.

2. Dec 2, 2004

### 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....)

3. Dec 2, 2004

### 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.

4. Dec 2, 2004

### 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.

5. Dec 2, 2004

### 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.

6. Dec 2, 2004

### 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.

7. Dec 2, 2004

### 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.

8. Dec 2, 2004

### 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 $|x+1|\geq |x|-|1|$.

9. Dec 2, 2004

### quasar987

Right! I did that part too mechanically. Thanks!