$f:[a, \infty ) \to \mathbb{R}$, $g:[a, \infty ) \to \mathbb{R}$ are functions such that $\lim_{x \to \infty} f(x) = \infty$, $\lim_{x \to \infty} g(x) = \infty$
If $\lim_{x \to \infty} (\frac{f(x)}{g(x)}) = 2$, then $\lim_{x \to \infty} (f(x)-g(x))= \infty$
What I gather from this affirmation is that if the limit of the division of two functions that diverge as x approaches infinity is a positive number bigger than 1 (I think) then the function $f(x)$ (the numerator) is considerably greater than $g(x)$ from a certain $x_{0}$ onward, so $f(x)-g(x)$ diverges. While I can think of a few examples and intuitively know this to be true, I still struggle to prove it. From my notes I understand that what I'm trying to prove is that: "$\forall M>0$, $\exists x_{0} \in \mathbb{R}$ such that $\forall x\geq x_{0}$ $(f(x)-g(x))>M$". However I can't seem to get to this definition; I believe I'm making a mistake in the sense that I'm probably forgetting to use some limit properties or I'm using my hypotheses incorrectly.
Using the definitions in my notes, the information I believe to have is that: "$\forall M>0$, $\exists x_{0} \in \mathbb{R}$ such that $\forall x\geq x_{0}$, $f(x)>M$" because f(x) diverges as x approaches infinity; "$\forall M>0$, $\exists x_{0} \in \mathbb{R}$ such that $\forall x\geq x_{0}$, $g(x)>M$" because g(x) diverges as x approaches infinity too; "$\forall \epsilon>0$, $\exists x_{0} \in \mathbb{R}$ such that $\forall x\geq x_{0}$, $|\frac{f(x)}{g(x)}-2|<\epsilon$" because $\frac{f(x)}{g(x)}$ converges to 2 as x approaches infinity.
Regardless, I'm stumped and would greatly appreciate some help.
Best Answer
If $x\in[a,\infty)$, then$$f(x)-g(x)=g(x)\left(\frac{f(x)}{g(x)}-1\right).\tag1$$So, given $M>0$, take $N>0$ such that, when $x>N$,$$g(x)>2M\text{ and }\frac{f(x)}{g(x)}>\frac32.$$Then$$x>N\implies g(x)>2\text{ and }\frac{f(x)}{g(x)}-1$$and so it follows from $(1)$ that$$x>N\implies f(x)-g(x)>M.$$