Logic for limits

Question

How $1^\infty$ is an indeterminate form?

Anything raised to the power of 1 is 1:p

I know that infinity is just a symbol to denote an unknown thing but still confused.

Answer 1

Infinity is not a real number, so "anything raised to the power of 1" isn't applicable (which, by the way, is completely backwards... I think you mean "1 raised to any power" where "any" means "any real number.").

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Now to see that we cannot just blindly assign a value to a limit of this form, notice that

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

while

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

and

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

and

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

As you can see, in each case the actual limit depends on how quickly the base tends to $1$ and how quickly the exponent tends towards infinity.

In some sense: if the base tends to $1$ quickly enough, the limit is $1$; if the exponent blows up too rapidly, the limit is infinite.

Answer 2

Consider the sequence: $$a_n = (1 + \frac{1}{n})^n$$

Now as $n \rightarrow \infty$, the inside of the expression goes to $1$ and the exponent goes to $\infty$. However, the limit of this sequence is $e$.

Answer 3

Note that for

• $f(x)\to 1$
• $g(x)\to \infty$

$$f(x)^{g(x)}=e^{g(x)\log(f(x))}$$

and

$$g(x)\log(f(x))=\infty\cdot 0$$

which is indeterminate.

Answer 4

If $a>0$ then $a^b=e^{b\ln(a)}$. So we may say (informally) that $$1^{\infty}=\exp(\infty\cdot\ln(1))=\exp(\infty\cdot 0)$$ which is indeterminate because of the product $\infty\cdot 0$.

As an explicit example, note that for $a>0$ the following limit, which is an indeterminate form $1^{\infty}$, depends on the value of $a$: $$\begin{align}\lim_{n\to +\infty }\left(1+\frac{1}{n^a}\right)^{n}&= \lim_{n\to +\infty }\exp\left(n\ln(1+1/n^a)\right)\\ &=\lim_{n\to +\infty }\exp\left(1/n^{a-1}\right)=\begin{cases} 1 &\text{if $a>1$,}\\ e&\text{if $a=1$,}\\ +\infty&\text{if $0<a<1$.} \end{cases}\end{align}$$