mathematical concept of something without any limit
Factsheet
Videos
What do you mean by infinity?
What is the symbol of infinity?
What is the additive property of infinity?
I know infinity is not a number so is this question undefined or something Can anyone explain it to me please
What $1^\infty$ is, or is not, is merely a matter of definition. Normally, one would only define $a^b$ for some specific class of pairs of $a,b$ - say $b$ - positive integer, $a$ - real number.
When extending the definition of exponentiation to more general pairs, the key thing people keep in mind is that various nice properties are preserved. For instance, for $ b$ - positive integer, you want to put $a^{-b} = \frac{1}{a^b}$ so that the rule $a^ba^c = a^{b+c}$ is preserved.
It may make sense in some context to speak of infinities in the context of limits, but this is usually more a rule of thumb than rigorous mathematics. This may be seen as extending the rule that $(a,b) \mapsto a^b$ is continuous (i.e. if $\lim_n a_n = a$ and $\lim_n b_n = b$, then $\lim_n a_n^{b_n} = a^b$) to allow for $b_n \to \infty$. For instance, you may risk saying that: $$\lim_{n} (2+\frac{1}{n})^n = 2^{\infty} = \infty$$ If you agree to use rules of this kind, you might be tempted to also say: $$\lim_{n} (1+\frac{1}{n})^n = 1^{\infty} = 1$$ but this would lead you astray, since in reality: $$\lim_{n} (1+\frac{1}{n})^n = e \neq 1$$ Thus, it is safer to leave $1^\infty$ undefined.
A more thorough discussion can be found on Wikipedia.
When your teacher talks about $0/0$ or $\infty/\infty$ or $1^\infty$ he/she's not talking about numbers, but about functions, more precisely about limits of functions.
It's just a convenient expression, but it should not be confused with computations on simple numbers (which $\infty$ isn't, by the way).
When $1^\infty$ is referred to, it is to mean the following situation: there are two functions $f$ and $g$ defined in a neighborhood of $c$, with the properties
$\lim\limits_{x\to c} f(x)=1$
$\lim\limits_{x\to c} g(x)=\infty$ (or $-\infty$)
(of course, $c$ can also be $\infty$ or $-\infty$).
Saying that $1^\infty$ is an indeterminate form is just a mnemonic way to say that you cannot compute
$$\lim_{x\to c}f(x)^{g(x)}$$
just by saying “the base goes to $1$, so the limit is $1$ because $1^t=1$”. Indeed this can be grossly wrong as the fundamental example
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$$
shows.
Why is that? It's easy if you always write $f(x)^{g(x)}$ as $\exp(g(x)\log f(x))$ and compute the limit of $g(x)\log f(x)$, then applying the properties of the exponential function.
In the case above we'd have
$\lim\limits_{x\to c} \log f(x)=0$
$\lim\limits_{x\to c} g(x)=\infty$ (or $-\infty$)
so the limit
$$\lim_{x\to c}g(x)\log f(x)$$
is in the other $\infty\cdot0$ indeterminate form (that you should know). Why is it “indeterminate”? Because we have many instances of that form where the limit is not predictable by simply doing a (nonsense) multiplication:
\begin{gather} \lim_{x\to 0+}x\cdot\frac{1}{x}=1\\ \lim_{x\to 0+}x^2\cdot\frac{1}{x}=0\\ \lim_{x\to 0+}x\cdot\frac{1}{x^2}=\infty \end{gather}