Regarding limits and $1^\infty$

Possible Duplicate: ¢ Why is $1^{\infty}$ considered to be an indeterminate form

I have some inquiries concerning limits and also the undefinability of $1^\infty$.

As an example, is $\lim_{x\to\infty}1^x$ uncertain? Why is it not $1$? Or do mathematicians, when claiming that $1^\infty$ is uncertain, in fact describe instances such as $lim_{x\to\infty} \left(1 + \frac{a}{x}\right)^x$ where despite the fact that at a first glace the outcome is $1$, this is in fact a grandfather clause and also it amounts to $e^a$?

2022-06-07 14:34:55
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Answers: 3

When individuals claim that $1^{\infty}$ is an indeterminate kind, what they suggest is that if $f(x)$ is a function such that $\lim_{x \to r} f(x) = 1$ and also $g(x)$ is a function such that $\lim_{x \to r} g(x) = \infty$, the value of $\lim_{x \to r} f(x)^{g(x)}$ is not distinctly established as a whole. It is established in the grandfather clause that $f(x) = 1$, in which instance the restriction is clearly simply $1$.

2022-06-07 14:55:04

By itself, $1^{\infty}$ does not suggest anything, other than the effects that you had an amount of the kind $a^b$, and also $\lim_{x\rightarrow\infty}a=1$ and also $\lim_{x\rightarrow\infty}b=\infty$. Relying on just how promptly and also from what instructions $a$ comes close to 1, and also just how promptly $b$ strategies $\infty$, $a^b$ can come close to any kind of non - adverse number, infinity, or absolutely nothing specifically in all.

2022-06-07 14:55:00

Look up indeterminate forms. It relies on just how you analyze $1^{\infty}$

The indeterminate kind $1^{\infty} = \lim_{n \rightarrow \infty} f(n)^{g(n)}$ where $\lim_{n \rightarrow \infty} f(n) = 1$ and also $\lim_{n \rightarrow \infty} g(n) = \infty$

For instance, every one of the adhering to limits are of the kind $1^{\infty}$, yet they all review to various numbers

$$\lim_{n\rightarrow \infty} 1^n = 1$$

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

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

$$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{\log(n)} \right)^n = 0$$

It is an inquiry of which one often tends much faster whether $f(n) \rightarrow 1$ or $g(n) \rightarrow \infty$.

When you get a limited number which is non - absolutely no and also not one as the restriction, there remains in some feeling an equilibrium in between the price at which $f(n) \rightarrow 1$ and also $g(n) \rightarrow \infty$.

When you get infinity or absolutely no as the restriction, $g(n) \rightarrow \infty$ faster than $f(n) \rightarrow 1$.

When you get $1$ as the restriction, $f(n) \rightarrow 1$ faster than $g(n) \rightarrow \infty$.

2022-06-07 14:54:51