A basic doubt on rationals approximating reals

Suppose I want to define $2^{\sqrt2}$. Now there can be more than one sequence $\{x_n\}$ of rationals which converges to $\sqrt2$. Are there any relations between those sequences ? What is the guarantee that all of $\{2^{x_n}\}$ converge to the same limit ? Only a hint is enough.

• Think about continuity. – Martin Citoler Sep 15 '13 at 17:51
• An easier to manage concept of power (with irrational exponents) is to define $e^x$ by its Taylos series and build from that: $2^x=e^{x\ln 2}$. If you have been hit with this exercise, and have two sequences of rationals $(x_n)$ and $(y_n)$, then fix a tolerance $\epsilon>0$. From some index $\ell$ onwards we have $|x_n-y_n|<\epsilon$ whenever $n>\ell$. Then the ratio of $2^{x_n}/2^{y_n}$ (or the other way round) is within $2^\epsilon-1$ from $1$. This gives you a useful bound for the difference $|2^{x_n}-2^{y_n}|$. – Jyrki Lahtonen Sep 15 '13 at 17:58
• @MartinCitoler:understood, thanks. – aaaaaa Sep 15 '13 at 17:58

For every $x$ such that $|x-\sqrt2|\leqslant1$, $|2^x-2^{\sqrt2}|\leqslant2^{\sqrt2}\cdot|x-\sqrt2|$.

Thus, if $x_n\to\sqrt2$ and $y_n\to\sqrt2$, $2^{x_n}-2^{y_n}\to0$, $2^{x_n}\to2^{\sqrt2}$ and $2^{y_n}\to2^{\sqrt2}$.

Let $(x_n)$ and $(y_n)$ be two such sequences. Consider the ratio $\dfrac{2^{x_n}}{2^{y_n}}=2^{x_n-y_n}$.

Let $a_n=x_n-y_n$. Then the sequence $(a_n)$ has limit $0$. Show that $\lim_{n\to\infty}2^{a_n}=1$.