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.

  • 3
    $\begingroup$ Think about continuity. $\endgroup$ – Martin Citoler Sep 15 '13 at 17:51
  • 1
    $\begingroup$ 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 $|2^{x_n}-2^{y_n}|$. $\endgroup$ – Jyrki Lahtonen Sep 15 '13 at 17:58
  • $\begingroup$ @MartinCitoler:understood, thanks. $\endgroup$ – 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 $\lim_{n\to\infty}2^{a_n}=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.