# help solving for delta in epsilon delta proof

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

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

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

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$$\lvert x-9\rvert<\delta\implies\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}<\frac{\lvert9-x\rvert}{30}<\varepsilon.$$

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

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

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

$$\epsilon >0$$ 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;

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

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

$$|\dfrac{1}{√x}-1/3| <$$

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