2
$\begingroup$

$$\lim_{x\to 9} \frac{1}{\sqrt{x}} = \frac{1}{3} $$

so the two statements are

$$0<|x-9|<\delta$$

$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$

I've tried to multiply by the conjegate to get

$$\frac{\frac{1}{x}-\frac{1}{9}}{\sqrt{\frac{1}{x}}+\frac{1}{3}}<\epsilon$$

I'm unsure where to go from here to get that x-9 factor i need

$\endgroup$
  • $\begingroup$ You can multiply $1/x$ by $9$ and $1/9$ by $x$ to get the numerator in the form $(9-x)/(9x)$. $\endgroup$ – Math1000 Oct 21 at 19:24
4
$\begingroup$

Note that$$\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\left\lvert3-\sqrt x\right\rvert}{3\sqrt x}=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}.$$Now, if $\lvert x-9\rvert<5$, then $x>4$ and therefore $\sqrt x>2$. Also, $3+\sqrt x>5$. So $3\sqrt x\left(3+\sqrt x\right)>30$. Therefore, if you take $\delta=\min\left\{5,30\varepsilon\right\}$, then$$\lvert x-9\rvert<\delta\implies\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}<\frac{\lvert9-x\rvert}{30}<\varepsilon.$$

$\endgroup$
  • $\begingroup$ is there any particular reason you set delta to 5? or was it just to make the roots look cleaner than the guy below? $\endgroup$ – Mohammad Ali Oct 21 at 19:36
  • $\begingroup$ You got it right: it was just to make the roots look cleaner. $\endgroup$ – José Carlos Santos Oct 21 at 19:38
4
$\begingroup$

We have that

$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{\sqrt{x}-3}{3\sqrt{x}}\right|=\left|\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{3\sqrt{x}(\sqrt{x}+3)}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|$$

then assuming wlog $|x-9|<1 \implies 3\sqrt{x}(\sqrt{x}+3)\ge 3\sqrt{8}(\sqrt{8}+3)$ therefore

$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|\le\frac{|x-9|}{ 3\sqrt{8}(\sqrt{8}+3)}$$

and it suffices to assume

$$\delta<3\sqrt{8}(\sqrt{8}+3)\epsilon $$

to have

$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$

$\endgroup$
2
$\begingroup$

$\epsilon >0$ be given.

$ |\dfrac{1}{√x}-1/3|=|\dfrac{3-√x}{3√x}|=$

$|\dfrac{9-x}{(3+√x)3√x}|\lt |\dfrac{9-x}{x}|$;

Consider $|x-9|<1$, then $8<x<10$;

Choose $\delta = \min(1, 8\epsilon)$;

Then $|x-9|<\delta$ implies

$|\dfrac{1}{√x}-1/3| <$

$|\dfrac{9-x}{x}| <(1/8)\delta \lt \epsilon.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.