Why a number/infinitesimal+ is equal to infinity? Why a number/infinitesimal- is equal to -infinity? If a real number is 5, and a infinitesimal positive is 0+, is 0.1 for example, no? I got this doubt while I was doing a limit: $$\lim_{x \to - 2^-}\frac{(x-5)}{(x-2)(x+2)}$$ $$\lim_{x \to -2^-}\frac{(-2-5)}{(-2-2)(x+2)}$$ $$\lim_{x \to -2^-}\frac{-7}{(x-2)0^+}$$ $$\lim_{x \to -2^-}\frac{-7}{((-4)(0^+)}$$ $$\lim_{x \to -2^-}\frac{(-7)}{-0^+}$$ Why is equal to infinity??
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Sorry but your question "Why a number/infinitesimal- is equal to -infinity?" makes no sense. – Oct 23 '20 at 17:28
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$\lim_{x \to 2^-}$ does not mean the limit as $x$ approaches negative two. It means the limit as $x$ approaches two from the left. Otherwise, a positive or negative infinitesimal can be treated as a positive or negative number. – David Diaz Oct 23 '20 at 17:31
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yeah is a typo, i edited – Oct 23 '20 at 17:37
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Do you understand the difference between one-sided and two-sided limits? – Noah Schweber Oct 23 '20 at 17:37
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yes, was a typo – Oct 23 '20 at 17:39
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number/infnitesimal means n/0^+ or n/0^- – Oct 23 '20 at 17:40
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limit of x->-2^- of 7/-0^+ is equal to infinty, why? – Oct 23 '20 at 17:42
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2Do you understand that if $h$ is very small and positive then $7\over h$ is very large and positive? – Noah Schweber Oct 23 '20 at 17:43
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Noah’s excellent answer should guide you when you rework the problem. In case its not yet clear (and if I understand your shorthand correctly), $\lim_{x \to -2^{-}}(x+2) \not = 0^{+}$. – David Diaz Oct 23 '20 at 19:50
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From the comments below, it seems the point of confusion is how to understand $0^+$. The answer is simply:
Don't use it (yet).
"$0^+$" is a shorthand used to abbreviate limit calculations, but it shouldn't be used until those precise calculations are already well-understood. In particular, rather than think about "${-7\over -0^+}$" you should force yourself to think explicitly in terms of $$\lim_{h\rightarrow 0^+}{-7\over -h}.$$
- Wait, isn't there a "$0^+$" in that expression already? Well, not really: it's not "the limit as $h$ approaches the number $0^+$" but rather "the limit as $h$ approaches the number $0$ from the right." Actually way back when I was learning calculus I was quite annoyed by this notation, and argued that it should be "$h\rightarrow^+0$" instead since "${}^+$" is modifying the way the limit is taken, not the value it's taken at. But anyways.
Now plugging in numbers does not a valid calculation make, but it may nonetheless be a useful reality check. Take for example $h={1\over 100000000}$. Then $${-7\over -h}={-7\over-({1\over 100000000})}={7\over({1\over 100000000})}=700000000.$$ This should suggest that if $h$ is really small and positive then ${-7\over -h}$ is really big and positive, so we should expect $\lim_{h\rightarrow 0^+}{-7\over -h}=+\infty$.
(Note that we could have simplified things right from the start and cancelled the minus signs: it's probably easier to think about $\lim_{h\rightarrow 0^+}{7\over h}$, and that's the same thing.)
In general, you need to force yourself to go slowly with limits and use the definitions carefully until you've mastered them. There are lots of places where going too fast will get you in trouble. Another common stumbling block (I've seen it frequently with students) is about the converses of the limit laws. For example, I've seen many students argue that $\lim_{x\rightarrow a}[f(x)+g(x)]$ is undefined whenever both $\lim_{x\rightarrow a}f(x)$ and $\lim_{x\rightarrow a}g(x)$ are undefined, but of course that's incorrect: consider $$f(x)={1\over x},\quad g(x)=-{1\over x}\quad\implies \quad f(x)+g(x)={1\over x}-{1\over x}.$$
Noah Schweber
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@MarkosAndres "$0^+$" isn't actually a thing. We need to talk in terms of limits. If you understand why ${1\over h}$ is very large and positive when $h$ is very small and positive, then you understand that $\lim_{h\rightarrow 0^+}{1\over h}=+\infty$. – Noah Schweber Oct 23 '20 at 17:45
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Notice that in your example you don't have $0^+$ in the denominator but rather $-0^+$. – Noah Schweber Oct 23 '20 at 17:46
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@MarkosAndres Again that's meaningless; you need to phrase things in terms of limits. In fact $\lim_{h\rightarrow 0^+}{-7\over h}=-\infty$, not $+\infty$; but the expression you're considering in the OP is (equivalent to) $\lim_{h\rightarrow 0^+}{-7\over -h}$, which is not the same (and is in fact $+\infty$). Look at the denominators! – Noah Schweber Oct 23 '20 at 17:47
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ah is 7/0.9, 7/0.99, 7/0.999 and is more big, that the limit is equal to infinity? – Oct 23 '20 at 17:51
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@MarkosAndres No, you want to think about $7\over 0.1$, $7\over 0.01$, $7\over 0.001$, ... The sequence $0.9, 0.99, 0.999,...$ goes to $1$, not to $0$. – Noah Schweber Oct 23 '20 at 17:52
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yeah, error of concentration sorry. now i understand. I was looking at 0 ^ + as a fixed real number not as a number that tends towards a value – Oct 23 '20 at 17:54
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@MarkosAndres First of all, I'll say again: don't use $0^+$ as an actual thing. Just talk about actual numbers and limits; this will avoid confusion. To understand what the expression $$\lim_{x\rightarrow-2^+}{x-5\over (x+2)(x-2)}$$ should be, think about the value of ${h-5\over (h+2)(h-2)}$ for $h$ "just a little bit bigger" than $-2$. What can you say about that? – Noah Schweber Oct 23 '20 at 18:06
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(I just noticed that at some point you edited the question to change from $\rightarrow 2^+$ to $\rightarrow -2^+$ btw.) – Noah Schweber Oct 23 '20 at 18:06
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ok i understand, but when a must calculate limit of fuction in a point that is not allow in the domain, i must always check the left side and right side limit? – Oct 23 '20 at 18:34
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@MarkosAndres Yes (unless the question asks about only one of the two sides). – Noah Schweber Oct 23 '20 at 18:40
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One of operation of the limits is: lim x->a (f(x)-g(x))= limx->a f(x)- limx->a g(x). Then why the limit x->1 (1/(x^2-1))-(2(x^4-1))=1/2 if limx->a of 1/(x^2-1) dont exist and lim of x->a 2/(x^4-1) dont exist? – Oct 23 '20 at 18:46
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@MarkosAndres Don't try to blindly guess how the limit laws work; look at the actual definition of the limit. Consider the following simpler example: $f(x)={1\over x}, g(x)=-{1\over x}$, $(f+g)(x)={1\over x}-{1\over x}$. Of course neither $\lim_{x\rightarrow 0}f(x)$ and $\lim_{x\rightarrow 0}g(x)$ exists but $\lim_{x\rightarrow 0}(f+g)(x)$ does exist and is $0$. "Splitting up" a limit doesn't always work: we can only go one way (from the individual summands to the sum) and even then only sometimes (think about situations where one summand goes to $\infty$ and the other goes to $-\infty$). – Noah Schweber Oct 23 '20 at 18:52
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therefore none the non-existence of f (x) and g (x) does not imply the non-existence of f (x) -g (x)? Does it apply to all properties? – Oct 23 '20 at 18:56
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@MarkosAndres Try to think of similar examples! Figuring out the details of the limit rules is exactly how you'll master the concept. And at this point we're getting a bit far afield from the original question, so I suggest asking a new question instead of continuing this comment thread. – Noah Schweber Oct 23 '20 at 19:03
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yeah i ask always question but i have a question not math-related, is it serious if I don't understand the formal definitions right away? – Oct 23 '20 at 19:28
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@MarkosAndres No - it's rare that someone understands the formal definitions right away. But the right response is to slow down and focus on mastering them rather than leaping ahead and trying to learn the general rules that let you manipulate them quickly. – Noah Schweber Oct 23 '20 at 19:29
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OK thank you, you advice study the formal definition and after the intuitive definition, or the other way around? – Oct 23 '20 at 23:20