2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ We can have $g(x)=0$ for $x\ge a$ if $a<0$ $\endgroup$ – hamam_Abdallah May 10 at 0:47
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos May 10 at 6:17
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.