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If $\lim_{ x\to0} f(x) = \infty$ and $\lim_{ x\to0} g(x) = \infty$, then $\lim_{ x\to0} [f(x) − g(x)] = 0$. True or False??

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almost always false – Pocho la pantera Sep 13 '13 at 19:22
indeed a good example of why you should not even use the notaion lim<whatever> "equals" <infintiy> – george Sep 13 '13 at 20:31
@george: It's even better to keep this extremely useful notation, and simply learn how to do arithmetic correctly with extended real numbers. – Hurkyl Sep 15 '13 at 3:14
@george as Hurkyl says, $\infty-\infty$ is not defined in extended real numbers, so there is no problem with using this. – chubakueno Sep 15 '13 at 3:15

For example $$ f(x)=\frac2{x}, \ g(x)=\frac1{x}. $$

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Let $f(x)=\dfrac{3m}{\sin^2x}$ and $g(x)=\dfrac{3m}{x^2}$ $$\lim_{x\to0}[f(x)-g(x)]\neq0 \quad,\forall m\in \mathbb N\cap\{m\ge 1\}$$

Edit:$$\lim_{x\to0}[f(x)-g(x)]=\lim_{x\to0}\left[\dfrac{3m}{\sin^2x} -\dfrac{3m}{x^2}\right]=\lim_{x\to0}\left[\dfrac{mx^4}{x^4}\right]$$

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Why do you have the condition $m\geq 4$ (or even $m\in\mathbb N$)? – Jonas Meyer Sep 13 '13 at 19:38
@Jonas Meyer : because $$\lim_{x\to0}[f(x)-g(x)]=\lim_{x\to0}[\dfrac{3m}{\sin^2x} -\dfrac{3m}{x^2}]=\lim_{x\to0}[\dfrac{mx^4}{x^4}]$$ – Maisam Hedyelloo Sep 13 '13 at 19:50
Maisam: I don't understand how your comment is relevant to my question. I see that you have now replaced the condition $m\geq 4$ with $m\geq 1$, but it doesn't matter whether $m$ is a natural number. The limit is $m$, so the only condition needed to not get $0$ is $m\neq 0$. I like the example. – Jonas Meyer Sep 13 '13 at 20:10

\begin{align} {\rm f}\left(x\right) = {1 \over x}\,, \quad {\rm g}\left(x\right) = {1 \over x}\,; & \qquad\qquad{\large\mbox{( TRUE )}} \\[3mm] {\rm f}\left(x\right) = {2 \over x}\,, \quad {\rm g}\left(x\right) = {1 \over x}\,; & \qquad\qquad{\large\mbox{( FALSE )}} \end{align}

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False. Let $f(x)=\dfrac1{x^2}$ and $g(x)=\dfrac1{x^2} + 1$.

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Blurry, as the examples have shown you, the essense here is that you cannot work with infinity as if it is a number. Infinity is an entity with which you cannot calculate in a normal fashion as with ordinary numbers. Just like saying infinity divided by infinity is equal to 1. Very often that is not true either. In case of infinity - infinity, usually making one term (most of the time fractional) out of them is the way to go. Feel free to post an actual limit problem and we will help you.

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