A proposition I would like to prove:
Proposition. Let $(a_i)_{i=1}^{\infty}$ be a convergent sequence of real numbers (converges to $L$) and let $x > 0$ be a real number. Then $$\lim_{n \rightarrow \infty}x^{a_n} = x^{L}$$
My attempt:
Note that if there is some sequence of rational numbers $(q_i)_{i=1}^{\infty}$ that converges to $L$, then we would have $\lim_{n \rightarrow \infty}x^{q_n} = x^{L}$ by definition of real exponentiation.
With this in mind, we will find sequences of rational numbers $(q_i)_{i=1}^{\infty}$,$(p_i)_{i=1}^{\infty}$ (both converge to $L$) such that $q_n \leq a_n \leq p_n$ for all $n$.
For positive natural number $n$, define $S_n := \{q \in \mathbb Q\mid a_n - 1/n \leq q_n \leq a_n \}$ and by using axiom of choice, define a sequence $(q_i)_{i=1}^{\infty}$ such that $q_n \in S_n$ for all $n \geq 1$. Since $a_n - 1/n \leq q_n \leq a_n$, using squeeze test one can see that $q_n \rightarrow L$ as $n \rightarrow \infty$.
Similarly we define $P_n := \{p \in \mathbb Q\mid a_n \leq p_n \leq a_n +1/n\}$ for all positive natural numbers $n$. Now we define sequence $(p_i)_{i=1}^{\infty}$ such that $p_n \in P_n$ for all $n \geq 1$. We can easily check that the sequence is convergent to $L$.
Now suppose that $x > 1$. Since $q_n \leq a_n \leq p_n$, we have $x^{q_n} \leq x^{a_n} \leq x^{p_n}$. By letting $n \rightarrow \infty$ and using squeeze test, we can see that $x^{a_n} \rightarrow x^L$ as $n \rightarrow \infty$.
Now suppose that $x < 1$. In this case we have $x^{q_n} \geq x^{a_n} \geq x^{p_n}$. Using squeeze test one more time, we see that the $(x^{a_n})_{n=1}^{\infty}$ is convergent to $x^L$.
$\Box$
Is it correct?