$f(x) =\frac{1}{x^{\alpha}}$ continuity and uniform continuity

Question

Let $f(x) =\frac{1}{x^{\alpha}}$where $x \in (0, 1)$ and $0<\alpha \le 1$. Is $f$ continuous? What about uniform continuity? Justify.

I know this function is continuous in the given domain for all such $\alpha$ but how do I show it rigourously?

For uniform continuity I need to show for any $\epsilon >0$ $ \exists \delta >0 $ for which $\vert x-y \vert < \delta\implies \vert f(x) - f(y) \vert < \epsilon $

How do I do this ?

Answer 1

For $0<\alpha\leqslant1$, we have $$f(x) = x^{-\alpha} = \exp\left(-\alpha\log x \right), $$ so continuity follows from the composition of continuous functions.

However, given the Cauchy sequence $x_n=\frac1n$ in $(0,1)$, we have $$f(x_n) = \left(\frac1n\right)^{-\alpha} = n^\alpha\stackrel{n\to\infty}\longrightarrow\infty, $$ so $f$ is not Cauchy-continuous, and therefore not uniformly continuous. To see this more explicitly, fix $m$ and consider the behavior of $|f(x_n)-f(x_m)|$ as $n\to\infty$.