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Is $\dfrac{1}{\frac{0}{0}-1}$ an indeterminate form? I thought only $\dfrac 00,\,\dfrac\infty\infty$ and any form that can be represented in those two are indeterminate. Moreover, how do we know if a form is indeterminate?

P.S.

For people who says that this question doesn't reflect OP's effort, I came through it when calculating $$\lim_{t\rightarrow 0} \frac{1}{\frac{2\sin^2\frac{t}{2}\sin t}{(1-\cos t)(t-\sin t)}-1}.$$ I went to a Wikipedia page on indeterminate forms and there wasn't any mention of it.

Tianlalu
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mathnoob123
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    It definitely is indeterminate. $0/0$ is a meaningless array of symbols. – AlvinL Jul 14 '17 at 09:26
  • So wherever $0/0$ or $\infty/\infty$ arises no matter how many terms are along with it, it is always an indeterminate? – mathnoob123 Jul 14 '17 at 09:27
  • Absolutely. Division can also be viewed as multiplication with inverse. The zero element $0$ does not have an inverse, hence the expression $0/0$ makes no sense. As for infinites. We can say that $a/\infty$ is zero and get away with it, but $\infty /\infty$ is again meaningless. – AlvinL Jul 14 '17 at 09:28
  • No I am aware that $0/0$ is indeterminate and why it is indeterminate. What I am asking is if $0/0$ comes along with some other terms such as $0/0xy$ where $x$ and $y$ are functions then does the whole expression becomes indeterminate? – mathnoob123 Jul 14 '17 at 09:30
  • This is a limit question. Your $0/0$ situation arises from $\frac{f(x)}{g(x)}$ as $x$ tends to some value. The answer isn't clear cut. It depends on what the $x*y$ exactly is. The bottom line is your initial expression is in an indeterminate form when you substitute in the $t=0$. You must do something with the problematic fraction. – AlvinL Jul 14 '17 at 09:34
  • Can you tell me some examples in which the expression $0/0xy$ ceases to be indeterminate? (Apologies for the ambiguity in my question, I hope you get what I am trying to convey) – mathnoob123 Jul 14 '17 at 09:37
  • I wrongly expressed myself. $0/0$ is always indeterminate, no matter what. You arrived at this supposed $0/0$ because you tried substituting in $t=0$. I meant that sometimes the expression $\frac{f(x)}{g(x)} * a(x) * b(x) $ can be reduced or manipulated in some way such that substitution no longer yields something indeterminate. – AlvinL Jul 14 '17 at 09:41
  • Okay thank you. But if it doesn't yield a reduced form then it is indeterminate right? The whole expression? – mathnoob123 Jul 14 '17 at 09:43
  • Now that is much more difficult to answer. There are many ways to manipulate an expression. Your ultimate question should, however, be whether the initial limit exists or not. The "indeterminate forms" are a convinient language we invented to say "direct substitution" didn't work (when dealing with limit calculations) – AlvinL Jul 14 '17 at 09:47
  • No I am concerned with the use of L Hospital Rule. Since they only work on indeterminate forms it's essential to know whether the expression is indeterminate or not. – mathnoob123 Jul 14 '17 at 09:48
  • Yes, indeed, that is one of your options. How would you manipulate the expression under the limit to apply L'Hopital's rule, though? – AlvinL Jul 14 '17 at 09:51
  • That's through differentiation. But to use that first we have to first ensure the expression is completely indeterminate. So that why I was concerned with $0/0*xy$ example. – mathnoob123 Jul 14 '17 at 09:52
  • That is confusing terminology. I have given you guidelines how to apply the rule. – AlvinL Jul 14 '17 at 10:04
  • Okay good thanks that was clear. Just one question can f(t) be a constant and g(t) be a 0/0 for this rule to apply ? – mathnoob123 Jul 14 '17 at 10:05
  • You must do something to the initial expression. Something about common denominators and such.. – AlvinL Jul 14 '17 at 10:07
  • Hmm okay but what if this seems to be impossible (I am not saying that would be impossible but the person cannot find out how to reduce the expression). Can he directly apply the l hospital rule or would that be wrong? – mathnoob123 Jul 14 '17 at 10:09
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    Oh okay that was very helpful. Thank you very much. – mathnoob123 Jul 14 '17 at 10:12
  • That limit seem to be 0. – Takahiro Waki Jul 14 '17 at 10:30

1 Answers1

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Let us consider the main purpose of this topic to solve the limit and discuss the method of solving it.

As I mentioned in the comments, the use of L'Hopital's rule is not determined by some or other "indeterminate forms". The theorem has very specific assumptions. Furthermore, I do not know what it means to be "completely indeterminate" and how that would specify the condition of being "indeterminate".

Suppose we needed to find $$\lim _{t\to 0} \frac{f(t)}{g(t)} $$ If $\lim _{t\to 0} f(t) = \lim_{t\to 0} g(t) =0$ OR $\lim_{t\to 0} f(t) =\pm\infty$ and $\lim _{t\to 0} g(t) =\pm\infty$, then $$\lim _{t\to 0} \frac{f(t)}{g(t)} = \lim _{t\to 0} \frac{f'(t)}{g'(t)} $$ Of course, $f$ and $g$ must be differentiable around $t=0$, but that is not an obstacle for this particular problem.

AlvinL
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