Prove that $\frac{1}{\sqrt{x}}$ is continuous on $(0,\infty)$.

Question

ATTEMPT (through the $\epsilon$-$\delta$ method)

Let $\epsilon$ > 0. Want $\delta > 0$ such that for every $y\in (0,\infty)$ we have

$$|x-y|<\delta \implies |f(x)-f(y)| < \epsilon.$$

Noting that

$$\bigg|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}\bigg| = \bigg|\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\bigg|=\frac{|x-y|}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}$$

Don't know a reasonable $\delta$ to pick. Help would be appreciated.

Answer 1

Correct me if wrong :

$x \in (0,\infty)$:

$\dfrac {|x-x_0|}{\sqrt{xx_0}(√x+√x_0)} \lt$

$\dfrac{|x-x_0|}{x√x_0}.$

Consider:

$|x-x_0|< x_0/2.$

Then $-x_0/2 +x_0 < x < x_0/2 +x_0$,

or $x_0/2 <x.$

Let $\epsilon >0$ be given.

Choose $\delta < \min (x_0/2, (x_0^{3/2}/2)\epsilon)$.

Then $|x-x_0| < \delta$ implies

$|1/√x-1/√x_0| < \dfrac {|x-x_0|}{x√x_0} \lt$

$\dfrac{2\delta}{x_0^{3/2}} \lt \epsilon.$

Answer 2

hint: as pointed out by the comment above, you should fix a number, say $a$ and you show continuous at $a$. All you got to do is replace your $y$ by $a$ in your lines above. Plus you may assume $|x-a| < \frac{a}{2}$. So $ x - a > -\frac{a}{2}$, and $x > \dfrac{a}{2}$. Thus $\sqrt{xa}(\sqrt{x}+\sqrt{a})>\dfrac{a\sqrt{a}}{2}$. Can you continue to select the suitable $\delta$ for this?