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I think that the following statement is true;

If $\lim_{\{f(x),g(x)\}\to\infty}\frac{f(x)}{g(x)}=1$, then $\lim_{\{f(x),g(x)\}\to\infty}f(x)-g(x)=0$.

But I haven't learned the rools of limits yet, so I don't know if it is.

Can anyone make me sure that it is true, or tell me that it's not...?

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    $\begingroup$ What does the limit as $f(x)$ and $g(x)$ go to infinity mean? What if they go to infinity at multiple points? Do you mean $\lim_{x\to\infty}\left(\frac{}{}\right)$? $\endgroup$
    – Sophie
    Commented Nov 17, 2016 at 16:11
  • $\begingroup$ look at $f(x) =x, g(x)= x+1$ $\endgroup$
    – Thomas
    Commented Nov 17, 2016 at 16:11
  • $\begingroup$ @Thomas Thank you, I didn't think of that... It's simple. $\endgroup$
    – 76david76
    Commented Nov 17, 2016 at 17:12
  • $\begingroup$ It's worth noting the converse isn't true either. For example, $f(x) = \frac{1}{2x}$, $g(x) = \frac 1x$. $\endgroup$ Commented Nov 22, 2016 at 22:13

2 Answers 2

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Consider $f(x) = x+1$ and $g(x)=x+2$. Or perhaps even better is $f(x) = g(x)+1$

What is your limit for $\frac{f(x)}{g(x)}$?

What is your limit for $f(x)-g(x)$?

Edit: In fact the limit can be $\infty$ if one sets say $f(x) = g(x)+\ln(g(x))$.

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  • $\begingroup$ Thank you, it's simple. Maybe can I ask this, will the limit $\lim f(x)-g(x)$ exist? or it can be infinity? $\endgroup$
    – 76david76
    Commented Nov 17, 2016 at 17:17
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    $\begingroup$ I know for certain you can make the limit be impossible to resolve by making the difference say $\cos(x)$. You could make the limit infinite by adding in say $\ln(x)$ or $\ln(g(x))$ where the limit will be zero for the proportion, but infinite on its own. $\endgroup$ Commented Nov 17, 2016 at 17:50
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Hint:

Let $f(x)=g(x)+h(x)$, where $\lim_{x\to \infty}\frac{h(x)}{g(x)}=0$ and $\lim_{x\to \infty}h(x) \ne 0$.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Commented Dec 2, 2016 at 14:38

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