Prove $f(x) = x^p$, $0<p<1$ is continuous for $x>0$

Question

I know how to prove specific cases like $f(x) = \sqrt{x}$ is continuous via the $\delta-\epsilon$ definition, but I'm not sure how to pick the $\delta$ when $p$ is not specified. Any help would be appreciated!

-EDIT- my attempt that was going nowhere:

WLOG let $x > x_0$, for any $\epsilon > 0$ we want to find $\delta$ for all $x_0 > 0$ such that $x - x_0 < \delta \Rightarrow x^p - x_0^p < \epsilon$.

So $x - x_0 < \delta \Rightarrow x^p < x_0^p(\delta / x_0 + 1)^p$

Take $\delta = x_0 \epsilon ^{1/p} - 1$ and substitute back, and we have $x^p < x_0^p \cdot \epsilon$, but this isn't what I want.

Answer 1

$f(x)=x^p$ is continuous at $x_0 \gt 0$

$\Longleftrightarrow$ $\lim_{x \rightarrow x_0} x^p = x_0^p$

$\Longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0, \forall 0 \lt |x - x_0| \lt \delta, |x^p - x_0^p| < \epsilon$

$\Longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0, \forall 0 \lt |\frac{x}{x_0} - 1| \lt \frac{\delta}{x_0}, |(\frac{x}{x_0})^p - 1| < \frac{\epsilon}{x_0^p}$

$\Longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0, \forall 0 \lt |x - 1| \lt \delta, |x^p - 1| < \epsilon$

$\Longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0, \forall 1-\delta \lt x \lt 1+\delta, x \neq 1, 1 - \epsilon \lt x^p < 1 + \epsilon$

$\Longleftrightarrow$ $\forall \epsilon > 0, \exists \delta > 0, \forall 1-\delta \lt x \lt 1+\delta, x \neq 1, (1-\epsilon)^\frac{1}{p} \lt x < (1+\epsilon)^\frac{1}{p}$

It is obvious to see that, we only need to put $\delta = min \{ 1 - (1-\epsilon)^\frac{1}{p}, (1+\epsilon)^\frac{1}{p} - 1 \} $ in order the rightmost part of the final step to be satisfied. Therefore, the proposition is proved.

Answer 2

Do you only want an argument using epsilon-delta? I mean you could write the function as $\exp(p\log x)$ and use the fact that composition of continuous functions is continuous.

Answer 3

Let us solve

$$x^p\pm\epsilon=(x\pm\delta)^p,$$ which gives

$$\pm\delta=\sqrt[p]{x^p\pm\epsilon}-x.$$

Then as the function is increasing, with

$$\delta=\min(-(\sqrt[p]{x^p-\epsilon}-x), \sqrt[p]{x^p+\epsilon}-x)$$ we achieve the desired bracketing. As this $\delta$ is defined for any $\epsilon$ (see the final note), continuity is ensured.

---

In fact, the function is convex, and the minimum value is always on the left, so we can use

$$\delta=-(\sqrt[p]{x^p-\epsilon}-x).$$

---

Final note:

When $\epsilon$ exceeds $x^p$, we have the power of a negative, which is not defined. In this case, we adjust the above expression by replacing $x^p-\epsilon$ by $0$.

$$\delta=\min(-(\sqrt[p]{\max(x^p-\epsilon,0)}-x), \sqrt[p]{x^p+\epsilon}-x).$$