Why is $1^{\infty}$ considered to be an indeterminate form

Question

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 their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

• The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$

My question is can anyone give me a nice explanation of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed.

Answer 1

Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities. For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.

$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$$

To expand on this some (and this thought process can be applied to other indeterminate forms, too), one way to think about it is that there's a race going on between the expression that's trying to go to 1 and the expression that's trying to go to $\infty$. If the expression that's going to 1 is in some sense faster, then the limit will evaluate to 1. If the expression that's going to $\infty$ is in some sense faster, then the limit will evaluate to $\infty$. If the two expressions are headed toward their respective values at essentially the same rate, then the two effects sort of cancel each other out and you get something strictly between 1 and $\infty$.

There are some other cases, too, like $$\lim_{n \to \infty} \left(1 - \frac{1}{\ln n}\right)^n = 0,$$ but this still has the expression going to $\infty$ "winning." Since $1 - \frac{1}{\ln n}$ is less than 1 (once $n > 1$), the exponentiation forces the limit to 0 rather than $\infty$.

Answer 2

Look at the logarithm.

More specifically, consider $f(x)^{g(x)}$ as $x \to \infty$, where $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to \infty} f(x) = 1$. (This is something of the form $1^\infty$.)

Now say $f(x) = e^{h(x)}$, so $h(x) = \log f(x)$. Then $$\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \log f(x) = \log \lim_{x \to \infty} f(x) = \log 1 = 0.$$

Then $$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} \exp \Big(g(x) \log f(x)\Big) = \exp \Big(\lim_{x \to \infty} g(x) \log f(x)\Big)$$ and since the limit of a product is the product of the limits, that's $$ \exp \left[ \left(\lim_{x \to \infty} g(x)\right) \cdot \left(\lim_{x \to \infty} \log f(x)\right)\right] $$ or

$$ \exp \left[ \left(\lim_{x \to \infty} g(x)\right) \cdot \left(\lim_{x \to \infty} h(x)\right)\right] $$

But the first limit of $g(x)$ is infinity, and the second of $h(x)$ is zero.

So the indeterminacy of $1^\infty$ follows directly from the indeterminacy of $\infty \cdot 0$.

(The indeterminacy of $\infty^0$ actually follows in the same way, by taking the factors in the other order.)

Answer 3

This is just one more consideration $1^\infty$ can be roughly rewritten as:

$1^{\frac 10}=\sqrt[0]{1}$

Now just think to the zeroth root of 1: every number raised to 0 is one so the zeroth root of 1 could be every number! This is why $1^\infty$ is an indeterminate form.

Answer 4

If you're confused about this part let me try to clarify:

$$1^\infty=1$$

$$\lim_{x\to\infty} 1^x=1$$

$$\lim_{x\to\infty} (1-\frac{1}{x} )^x=???$$

Only the last one is indeterminate. We can't be sure if the expression in parentheses goes to 1 "faster" than the exponent takes the entire expression to infinity. The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean. This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form. Hence, indeterminate form.

Answer 5

In fact, a better notation of this type of indeterminate form should be $(\rightarrow 1)^\infty$, where the right arrow means the number $1$ is the limit of the base function, not all of the value of the base function is literally $1$ (i.e. not the case like $\displaystyle\lim_{n\to\infty}1^{n}$).

This is the same when we write other indeterminate form, for example like $\frac{\to 0}{\to 0}$. If the nominator function is TRULY $0$, such like $\displaystyle\lim_{n\to\infty}\frac{0}{\frac{1}{n}}$, then it is not the indeterminate form $\frac{\to 0}{\to 0}$ that the calculus books are talking; it is definite form, $\displaystyle\lim_{n\to\infty}\frac{0}{\frac{1}{n}}=0$. To repeat, it seems to be, but NOT the indeterminate form, and we had better denoted it as $\frac{0}{\to 0}$.

Answer 6

"Indeterminate forms" are a vague concept and it is better to keep them "vague" rather than define them properly.

Limit evaluations are done on the basis of certain limit theorems which include the "algebra of limits" in particular. Theorems dealing with "algebra of limits" suffice to calculate limit of expressions which are composed of sub-expressions combined with $+, -, \times, /$ and the hope is that each sub-expression has a limit (perhaps calculated by expressing is as a combination of sub-sub-expressions) and then we use the algebra of limits to calculate limit of the expression by combining limits of sub-expressions via operations of $+, -, \times, /$.

These rules of "algebra of limits" however have two main limitations:

1) Limits of sub-expressions must exist (meaning they are finite, sorry I had to be explicit here to use the word "finite" as some textbooks treat limit $\infty$ also as "existing").

2) Rule dealing with division says that the limit of sub-expression in denominator should not be $0$.

"Indeterminate forms" were conceived to enumerate the cases where "algebra of limits" fails because of the above two limitations and for each of these cases certain other tactics / methodologies were developed. A classic case is expression of type $f(x)/g(x)$ where both $f(x), g(x)$ tend to $0$. Now to classify such cases the indeterminate form $0/0$ was invented. Similarly to deal with expressions of type $f(x)g(x)$ where $f(x) \to \infty$ and $g(x) \to 0$ the form $\infty\times 0$ was used. Also in each case where an indeterminate form was invented all the following options were possible: 1) limit exists, limit is $\pm \infty$ or there is oscillation. So classifying certain cases into "forms" did not guarantee the eventual limit, it only allowed us to use tactics and tools suitable to that form. Hence the word "indeterminate" was also added (we could not determine the limit by the form).

Cases like $f(x)g(x)$ where $f(x) \to \infty$ and $g(x) \to 1$ can't be handled by "algebra of limits" but these are not classified into indeterminate forms because there are theorems in this case which say that the resulting limit is $\infty$ so that the form is no longer "indeterminate" and we could perhaps classify such cases into "determinate forms" if we wanted.

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Coming to the form $1^{\infty}$ it is obvious that it is designed to handle expressions of type $\{f(x)\}^{g(x)}$ where $f(x) \to 1$ and $g(x) \to \infty$. In such case we can write the expression as $\exp\{g(x)\cdot\log f(x)\}$. By properties of $\log$ function if $f(x) \to 1$ then $\log f(x) \to 0$ and hence $\{g(x)\cdot\log f(x)\}$ is already an indeterminate form of the type $\infty\times 0$. Therefore $\exp\{g(x)\log f(x)\} = \{f(x)\}^{g(x)}$ also has to be considered as an indeterminate form and it is usually written in the notation $1^{\infty}$.

Answer 7

Here is some intuitive explanation, suitable also for non-mathematicians. Suppose that an imaginary basketball player has a probability $p = 0.999$ of making a free throw. The probability that he makes $10000$ free throws in a row is very small, the probability that he makes $100$ is high, and the probability that he makes $1000$ is approximately $e^{-1}$.

Answer 8

By continuity of the logarithm,

$$\log(\lim a^b)=\lim\log(a^b)=\lim(\log(a)\cdot b)$$

so that an indeterminate form $1^\infty$ is equivalent to a form $0\cdot\infty$.

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A concrete example:

$$\log\left(1.1^{10}\right)={0.9531017980\cdots}=\log(2.5937424601\cdots)$$ $$\log\left(1.01^{100}\right)={0.9950330853\cdots}=\log(2.7048138294\cdots)$$ $$\log\left(1.001^{1000}\right)={0.9995003330\cdots}=\log(2.7169239322\cdots)$$ $$\log\left(1.0001^{10000}\right)={0.9999500033\cdots}=\log(2.7181459268\cdots)$$ $$\cdots$$ $$\log\left(1^\infty\right)=1=\log(2.7182818284\cdots)$$

Answer 9

$$ \lim_{n\to\infty} \left( 1 + \frac a n\right)^n = e^a. $$

This limit depends on $a$. In other words, if the base approaches $1$ and the exponent approaches $\infty$, that's not enough to tell you what the limit is.