I've been thinking about this for a while and I'm 90% sure I'm not understanding this limit right, and I'd appreciate some feedback.
I have something that looks like
$$f(x) = g(x) - h(x)$$
$f(x)$, $g(x)$, and $h(x)$ are all just functions with the following properties. $g(x) > h(x)$ so that $f(x) > 0$. $h(x) > x$.
I want to find $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) - \lim_{x \to \infty} h(x)$$.
If I can show that $\lim_{x \to \infty} g(x) \to \infty$, and $\lim_{x \to \infty} h(x) \to \infty$, then this isn't terribly helpful.
This is the idea I had. Because I know that $h(x) > x$, then I can say
$$h(x) = (1+a)x$$
where $a$ is some positive number. And because I know that $g(x) > h(x)$, then I can say
$$g(x) = (1+a+b)x$$
where again, $b$ is some positive number.
Then if I put this all together, I get
$$f(x) = (1+a+b)x - (1+a)x = bx$$
And if I now take the limit, I get
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} bx \to \infty$$.
So that's an answer, but the issue is that even though I know $h(x) > x$, I could just as easily say
$$h(x) = x + a$$
and because $g(x) > h(x)$,
$$g(x) = x + a + b$$
and by this, we get
$$f(x) = x + a + b - x - a = b$$
Applying the limit here gives us
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty}b = b$$
Which of these would be correct? Is there a way to tell or do I need more information at this point? Is there a different way to pursue these limits?
Thanks all!
They are both equally wrong and need the same correction: $a$ and $b$ aren't necessarily constants
It is true that if $g(x)>h(x)>x$, then $h(x)=(1+a(x))x$ and $g(x)=(1+a(x)+b(x))x$ for strictly positive functions $a,b:\Bbb R^+\to\Bbb R^+$ (and the same can be said for the $h(x)=a(x)+x$, $g(x)=a(x)+b(x)+x$ case). However, this isn't really something that will help you with the general theory.
It could help in particular cases, depending on exactly what $g$ and $h$ are. But in such cases, I don't think you would think about it this abstractly. You wouldn't assign some $a$ and $b$ functions and use this result. You would just carry out the subtraction directly.