Convergence of power of a real sequence

Question

Let $x_n\in\mathbb{R}$ be a sequence converges to $x$ and $\alpha>0.$ Then does the sequence $x_n^{\alpha}$ converge to $x^\alpha$?

I know it is true for $\alpha\in\mathbb{N}$. What about if $\alpha$ will be a fraction?

Thanks.

Answer 1

If $x \geq 0$ this is true. If $x <0$ the powers are not even defined. Example: you cannot define $(-1)^\sqrt {2}$ as a real number.

Answer 2

When $x\ge 0$ all works fine, when $x<0$ it depends.

For example for $a>0$

• $x_n\to -a \implies (x_n)^{\frac13}\to -\sqrt[3]a$

bur for $\alpha=\frac12$ the expression becomes not well defined.

Answer 3

You can exchange a function and a limit when the function is continuous at the given point.

E.g. $$\lim_{x\to3}\cos(x)=\cos\left(\lim_{x\to3}x\right).$$

$$\lim_{x\to3}\sqrt x=\sqrt{\lim_{x\to3}x}.$$

This is because continuity allows you to replace a bounded neighborhood of the function evaluation by a bounded neighborhood of the argument.