Could the exponent be raised on the limit operator instead of the whole expression?

Question

Could we use $$\underset{x\to c}{\lim}^nf(x)$$ to denote $$\big(\lim_{x\to c}f(x)\big)^n$$

Answer 1

You could of course define it as such, but it is not 'official'. Note that $$(\lim_{x\to c}f(x))^n={\lim_{x\to c}}^n f(x)$$ is a function of $n$.

Answer 2

I don't think thats a good idea: $\lim^n$ looks like you apply the linear operator $\lim$ defined on some space of functions with codomain $\mathbb{R}\cup \{\infty,-\infty\}$ $n$ times, which is not what you want, because you have an actual multiplication $\lim\times \lim \times \dots \times\lim$ in mind.

Answer 3

This expression $$\underset{x\to c}{\lim}^nf(x) $$does not make sense.

On the other hand $$\big(\lim_{x\to c}f(x)\big)^n$$ means you take the limit first and raise the result to the $n^{th}$ power.