If $\lim_{n\to\infty} a_n=1$, then does $\lim_{n\to\infty}a_n^{3/2}=1$ also?

Question

I have tried rewriting $a_n^{3/2}=(a^{1/2}-1)(a_n+a_n^{1/2}+1)$, but can't seem to make any progress after that.

Would applying the triangle inequality help?

Answer 1

Hint: The function $x\mapsto x\sqrt x$ is continuous at 1 and use the sequential characterization of the limit.

Answer 2

Let $f(x)=x^{3\over 2}$, $\lim_n(f(a_n)=f(\lim_n a_n)$ since $f$ is continuous.

Answer 3

If $f$ is continuous then $\lim_{\infty} f(a_n)=f(\lim_\infty a_n)$. Using $f(x)=x^{3/2}$ you have the answer.

I think you can demonstrate it using epsilons and the triangular inequality.

Answer 4

A direct argument can go as follows (without invoking the continuity), for sufficiently large $n$, \begin{align} & |a_n^{3/2} - 1| = |\sqrt{a_n}^3 - 1^3| = |(\sqrt{a_n} - 1)(a_n + \sqrt{a_n} + 1)| \\ = & |a_n - 1| \frac{a_n + \sqrt{a_n} + 1}{\sqrt{a_n} + 1} \\ \leq & |a_n - 1| (1 + \sqrt{a_n}) \leq 2|a_n - 1|. \end{align}

You can then conclude with an $\varepsilon$-$N$ syntax.