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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10almost always false – Pocho la pantera Sep 13 '13 at 19:22
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1indeed a good example of why you should not even use the notaion lim
"equals" – 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. – Sep 15 '13 at 03:14
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@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 03:15
5 Answers
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No
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
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@Jonas Meyer : because $$\lim_{x\to0}[f(x)-g(x)]=\lim_{x\to0}[\dfrac{3m}{\sin^2x} -\dfrac{3m}{x^2}]=\lim_{x\to0}[\dfrac{mx^4}{x^4}]$$ – M.H Sep 13 '13 at 19:50
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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
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\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}
Felix Marin
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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.
imranfat
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