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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...?

76david76
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2 Answers2

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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 $f(x) = g(x)+\ln(g(x))$.

Kitter Catter
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  • Thank you, it's simple. Maybe can I ask this, will the limit $\lim f(x)-g(x)$ exist? or it can be infinity? – 76david76 Nov 17 '16 at 17:17
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    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.

     – Kitter Catter Nov 17 '16 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 $\lim_{x\to \infty}h(x) \ne 0$.

Mark Viola
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