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I just want to know, in calculating limits, when I do direct substitution, and it gives 3/0 instead of 0/0, does it mean for sure that the limit does not exist?

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$\lim_{x\to0} 3/x^2=+\infty$ –  enzotib Sep 15 '12 at 17:40
    
$3/0$ has no meaning, as there is no number which, multiplied by 0, gives you $3≠0$, and so division by zero is undefined. –  Babak S. Sep 15 '12 at 17:41
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It depends. In the case $\frac{1}{x^2}$, some people say that the limit does not exist, some say that the limit is $(+)\infty$. Depends on the conventions used in your course. But in any case, $\lim_{x\to 0} \frac{1}{x^3}$ does not exist. But we can talk about limits from the right and limits from the left. For example, one can write $\lim_{x\to 0-}\frac{1}{x}=-\infty$. But remember that always it is not the behaviour at $0$ that matters, but the behaviour near $0$. –  André Nicolas Sep 15 '12 at 17:59
    
given the definition of limit, how can someone says $1/x^2$ does not have a limit in $0$? –  enzotib Sep 15 '12 at 18:43
    
@enzotib: There are several definitions of limit. The basic definition deals with the sentence $\lim_{x\to a}f(x)=b$ where $a$ and $b$ are numbers. Of course "$+\infty$" is not a number. Extensions of the basic definition can be made, to define precisely the meaning of the phrase "$\lim_{x\to a}=\infty$." –  André Nicolas Sep 15 '12 at 19:25

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up vote 6 down vote accepted

Yes, it does mean that. Suppose $b_n\to 0$ and $c_n:=\frac{a_n}{b_n}\to L\in \mathbb R$. Then $$\lim_{n\to\infty} a_n = \lim_{n\to\infty}b_n c_n=0\cdot L=0.$$

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But we don't know if $L$, as you consider, is exist. The OP didn't mention this issue. He noted that the form he have is just $3/0$. :) –  Babak S. Sep 15 '12 at 17:56
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Well, Babak, the question was: If direct substitution gives us $3/0$, can we say that the limit exists? This suggests that the limit of the numerator is $3$ and the limit of the denominator is $0$, yes? Hagen's answer shows that if the limit were to exist (in the sense of being a real number), then we would have to have the limit of the numerator be $0$, not $3$. Thus, the limit can't exist (as the OP suspected and Hagen confirmed). –  Cameron Buie Sep 15 '12 at 18:20
    
@CameronBuie: Yes. Hagen is on the right way (+1). Thanks for lighting my mind. –  Babak S. Sep 15 '12 at 18:58

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