# If $f$ and $g$ diverge as x approaches infinity and $\lim_{x \to \infty} (\frac{f(x)}{g(x)}) = 2$, then $\lim_{x \to \infty} (f(x)-g(x))= \infty$

$$f:[a, \infty ) \to \mathbb{R}$$, $$g:[a, \infty ) \to \mathbb{R}$$ 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.

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$$x>N\implies f(x)-g(x)>M.$$

• We can have $g(x)=0$ for $x\ge a$ if $a<0$ – hamam_Abdallah May 10 at 0:47
• The sign of $a$ is irrelevant here. But yes, we can have $g(x)=0$. However, since $\lim_{x\to\infty}g(x)=\infty$, we have $g(x)>0$ if $x$ is large enough. – José Carlos Santos May 10 at 6:17

hint

For $$x$$ great enough, $$f(x)-g(x)=g(x)\Bigl(\frac{f(x)}{g(x)}-1\Bigr)$$

thus $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=2\; \implies$$

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

and $$f(x)-g(x) \sim g(x)\;\;\; (x\to \infty)$$

$$****************************$$ Given $$M>0$$, there exist $$A,B>a$$ such that $$x>A \implies g(x)>2M$$ $$x>B \implies (\frac{f(x)}{g(x)}-1>\frac 12$$

Let $$C=\max(A,B)$$.

then $$x>C\; \implies\; f(x)-g(x)>M$$