The title pretty much says it all.
If you have some n-dimensional die, and you roll the die n times, what is the probability you'll get at least one "roll match" - that you roll the number i on the ith roll?
My reasoning was:
$$\begin{align*} \text{ since } 1 - P(\text{no roll matches at all}) &= P(\text{ at least one match})\\[5pt] &= 1 - \left (\frac{(n-1)}{n} \right) ^n\\ \end{align*}$$
And then we take the limit as n approaches infinity
$$\begin{align*} &\lim_{n \to \infty} 1 - \left (\frac{(n-1)}{n} \right) ^n\\[5pt] &=\lim_{n \to \infty} 1 - \left (\frac{(\infty-1)}{\infty} \right) ^n\\[5pt] &=\lim_{n \to \infty} 1 -(1) ^\infty\\[5pt] &=\lim_{n \to \infty} 1 -(1)\\[5pt] &= 0\\[5pt] \end{align*}$$
And this kind of makes sense to me? It feels like as although we roll the die more, even though we're rolling it a very large amount of times, the probability of actually getting the right side approaches $0$.
But the answer on Slader (for my textbook) says this:
(b)
If we are dealing with an $n$-sided die then we have $P(C_i)=1/n$. Hence we compute
$$\lim_{n\to\infty} P(C_1 \cup \dots \cup C_n) = \lim_{n\to\infty} \bigg(1-\bigg(1-\frac{1}{n}\bigg)^n\bigg)=1-\lim_{n\to\infty}\bigg(1-\frac{1}{n}\bigg)^n,$$
where we note that $\lim_{n\to\infty}(1-1/n)^n=\lim_{n\to\infty}(1+\frac{-1}{n})^n$ is a famous limit from calculus/analysis, equalling the number $e^{-1}$ where $e \approx 2.718$. Thus our result is
$$1-e^{-1}\approx 0.632.$$
I would be grateful if someone could help me out! Thanks
In your last calculation, you seem to be using a "rule" that if $\lim_{n\to\infty} B(n) = 1$ and $\lim_{n\to\infty} E(n) = \infty$, then $\lim_{n\to\infty} B(n)^{E(n)} = 1$. This is an incorrect rule (and this example actually shows why).
This is actually an example of the "$1^\infty$ indeterminate form", which is called an indeterminate form precisely because knowing the limits of $B(n)$ and $E(n)$ is not enough information to determine the limit of $B(n)^{E^(n)}$. (This is in contrast to say $\lim_{n\to\infty} B(n) = 2$ and $\lim_{n\to\infty} E(n) = 5$ implying $\lim_{n\to\infty} B(n)^{E(n)} = 2^5$, where the individual limits are sufficient information to determine the complicated limit.) Indeterminate forms are a significant topic when studying limits, and various tools exist (such as l'Hôpital's rule) for figuring out their values.
By the way, the previous line of the calculation also invoked a false "rule", namely that if $\lim_{n\to\infty} N(n) = \infty$ and $\lim_{n\to\infty} D(n) = \infty$ then $\lim_{n\to\infty} \frac{N(n)}{D(n)} = 1$. This is another indeterminate form, the $\frac\infty\infty$ indeterminate form, and the example $\lim_{n\to\infty} \frac{2n-1}n$ shows that the answer is not always $1$. The OP happened to get the right answer here, but that was just good luck, not a valid step.