Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [indeterminate-forms]

If the expression obtained after any substitution during limit analysis does not give enough information to determine the original limit, it is known as an indeterminate form.

A mathematical expression can said to be indeterminate if it is not definitively or precisely determined. Certain forms of limits are said to be indeterminate when merely knowing the limiting behavior of individual parts of the expression is not sufficient to actually determine the overall limit.

Basically there are seven form of indeterminate 0 × ∞, 0/0, ∞0, ∞ − ∞, 1∞, 00 and ∞/∞.

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Zero to the zero power – is $0^0=1$?

Could someone provide me with a good explanation of why $0^0=1$? My train of thought: $x>0$ $0^x=0^{x-0}=0^x/0^0$, so $0^0=0^x/0^x=\,?$ Possible answers: $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$ $0^0=0^x/0^x=0/0$, which is undefined PS. I've read the…
Stas
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What is the result of $\infty - \infty$?

I would say $\infty - \infty=0$ because even though $\infty$ is an undetermined number, $\infty = \infty$. So $\infty-\infty=0$.
Pacerier
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Why is $1^{\infty}$ considered to be an indeterminate form

From Wikipedia: In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by…
user9413
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Why is $\infty \cdot 0$ not clearly equal to $0$?

I did a bit of math at school and it seems like an easy one - what am I missing? $$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$ $$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$ (i.e add $0$ to $0$ as many times…
Ashley Schroder
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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates. What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty}…
Eric Naslund
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I have learned that 1/0 is infinity, why isn't it minus infinity?

My brother was teaching me the basics of mathematics and we had some confusion about the positive and negative behavior of Zero. After reading a few post on this we came to know that it depends on the context of its use. Why do we take 1/0 as…
Rishav Mehta
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What's wrong with this reasoning that $\frac{\infty}{\infty}=0$?

$$\frac{n}{\infty} + \frac{n}{\infty} +\dots = \frac{\infty}{\infty}$$ You can always break up $\infty/\infty$ into the left hand side, where n is an arbitrary number. However, on the left hand side $\frac{n}{\infty}$ is always equal to $0$. Thus…
JobHunter69
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Is it wrong to tell children that $1/0 =$ NaN is incorrect, and should be $∞$?

I was on the tube and overheard a dad questioning his kids about maths. The children were probably about 11 or 12 years old. After several more mundane questions he asked his daughter what $1/0$ evaluated to. She stated that it had no answer. He…
leonigmig
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Is $0^\infty$ indeterminate?

Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate? Or is it only 1 raised to the infinity that is?
Skylion
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Why does simplifying a function give it another limit

I'm asked: $$\lim_{x\to 1} \frac{x^3 - 1}{x^2 + 2x -3}$$ This does obviously not evaluate since the denominator equals $0$. The solution is to: $$\lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+3)}$$ $$\lim_{x\to 1} \frac{x^2 + x + 1}{x +…
user347213
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Why does L'Hôpital's rule work for sequences?

Say, for the classic example, $\frac{\log(n)}{n}$, this sequence converges to zero, from applying L'Hôpital's rule. Why does it work in the discrete setting, when the rule is about differentiable functions? Is it because at infinity, it doesn't…
User001
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Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?

The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin…
user170231
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I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?

This is what I did for: $$\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$$ Check form: $\left({\infty \over \infty}\right)^{\infty}$. Apply L'Hospital's Rule to just $\lim \limits_{x \to \infty} {\left({3x-2…
Matt
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Why is infinity multiplied by zero considered zero here?

I watched an online video lecture by some professor and she was solving a convergence problem of the power series $$\sum_{n=1}^\infty n!x^n,$$ i.e., she was finding the values of $x$ for which this power series is convergent. She did the ratio test…
M.Samuel
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Why is $\frac{\ln\infty}{\infty}$ equal to $\frac\infty\infty$?

For me there is no intuitive explanation for this. Yes, i get that if you want to find the natural logarithm of a very high number (infinity) that the ln would be high too. But it does not grow nearly as fast as infinity itself, for example the…
Aron
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