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?
Finding if the limit does not exist
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calculus
limits
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0$\lim_{x\to0} 3/x^2=+\infty$ – 2012-09-15
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0$3/0$ has no meaning, as there is no number which, multiplied by 0, gives you $3≠0$, and so division by zero is undefined. – 2012-09-15
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1It 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** $0$. – 2012-09-15
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0given the definition of limit, how can someone says $1/x^2$ does not have a limit in $0$? – 2012-09-15
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0@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 "$\lim_{x\to a}=\infty$." – 2012-09-15
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0@Andre: Not every concept of number is so limited. Nor does everyone think that definition's significance to the overall concept of limits of real variables is any greater than simply being what's taught in calc 101. – 2012-09-16
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0@Hurkyl: Agreed. What I usually try to do is to gauge the academic level the questioner is at, and try to answer in a precise way roughly at that level, perhaps pushing *a little* beyond. – 2012-09-16
1 Answers
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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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0But 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 $3/0$. :) – 2012-09-15
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1Well, 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). – 2012-09-15
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0@CameronBuie: Yes. Hagen is on the right way (+1). Thanks for lighting my mind. – 2012-09-15