Let a be a positive real number $a>0$ and $\{ x_n\}$ be a sequence of rational numbers such that $\lim_{n \to \infty}x_n=0$ Show that $\lim_{n \to \infty} a^{x_n}=1$
My attempt
given = $e^{x_n \ln a} \to 1$
But any other alternative way? this was asked for 10 marks.
On
I suggest that, in order to prove your statement, we are not allowed to directly cite the continuity over $\mathbb{R}$ of the exponential function . That's becsuse, the continuity over $\mathbb{R}$ of the exponential function is just built on the continuity over $\mathbb{Q}$ . Thus, if we cite the the continuity over $\mathbb{R}$ of the exponential function to prove the present statement, this work is nothing but a circular argument. Here, I give a proper proof for that.
Let $a>1$. If $a=1$,that is trivial; if $0<a<1$, the proof may be similar. Just as you assumed, $x_1,x_2,\cdots,x_n,\cdots $ is a rational number sequence with the limit $0$. Thus, $$|x_n|<\frac{1}{n}$$ holds starting from some term. Hence $$-\frac{1}{n}<x_n <\frac{1}{n}.$$
We may readily obtain $$\left(\frac{1}{a}\right)^{1/n}<a^{x_n}<a^{1/n}.$$ But it's well-known that $x^{1/n} \to 1 (n \to \infty)$ for $x>0.$ Thus, by squeeze theorem,we have $$a^{x_n} \to 1.~~~(n \to \infty).$$
You can prove this at a much more elementary level - long before you learn about logarithms, the exponential function, and continuous functions.
If $a=1$ then the proposition is trivial. Suppose you can prove it for all $a>1$; then it is also proved for all $a < 1$ (still $a > 0$), because $a^{x_n} = \big(\frac 1 a\big)^{-x_n}$, $\frac 1 a > 1$ if $a < 1$, and ${-x_n} \to 0$ if $x_n \to 0$.
Now let's fix $a > 1$. First let's show that $a^{\frac 1 n} \to 1$. Write $w_n = a^{\frac 1 n} - 1$; $w_n > 0$ and $a = (1 + w_n)^n \ge 1 + n \dot w_n$ (elementary inequality which can be proven by induction), so $w_n \le \frac{a-1} n$, which proves that $w_n \to 0$ as $n \to \infty$, as claimed.
Then take an arbitrary sequence $x_n$ that converges to 0. For every $N \gg 0$, $x_n$ is between $-\frac 1 N$ and $\frac 1 N$ for all $n$ sufficiently large. Conclude by using the special case we just finished proving.