## eva12 one year ago 1)test review let f(x)=1/sqrt(x) and 0<c<1. Then f is continuous at x=c.proof

1. eva12

definition 55.a function f is said to be continuous at a point x if for every epsilon >0 there exists $\delta>0$such that |f(y)-f(x)| < epsilon for all y such that |y-x| $<\delta$

2. zepdrix

For $$\large\rm f(x)=\frac{1}{\sqrt{x}}$$ to be continuous at c means $$\large\rm \forall\epsilon>0~~ \exists\delta>0:(x\in(0,1),|x-c|<\delta)\implies \color{orangered}{|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{c}}|<\epsilon}$$ Let's fiddle around with that orange part a little bit, maybe we can get somewhere.

3. zepdrix

Getting a common denominator gives us: $$\large\rm \left\lvert\dfrac{\sqrt{c}-\sqrt{x}}{\sqrt{cx}}\right\rvert$$ since we're in absolute value, this is equivalent to$$\large\rm \dfrac{1}{\left\lvert\sqrt{cx}\right\rvert}\left\lvert\sqrt x-\sqrt c\right\rvert$$ which is still less than our $$\large\rm \epsilon$$.$\large\rm \dfrac{1}{\left\lvert\sqrt{cx}\right\rvert}\left\lvert\sqrt x-\sqrt c\right\rvert<\epsilon$Multiply the sqrt(cx) to the other side, and multiply both sides by the conjugate of what's in the absolute,$\large\rm \left\lvert\sqrt x-\sqrt c\right\rvert\left\lvert\sqrt x+\sqrt c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert$$\large\rm \left\lvert x-c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert$

4. zepdrix

I think we can drop the absolutes on the right safely... since it's all addition and square roots,$\large\rm \left\lvert x-c\right\rvert<\epsilon\left(\sqrt x+\sqrt c\right)\sqrt{cx}$

5. zepdrix

Actually, lemme color these a sec before I go any further,$\large\rm \left\lvert x-c\right\rvert<\epsilon\color{royalblue}{\left(\sqrt x+\sqrt c\right)}\color{green}{\sqrt{cx}}$ So we were able to, sort of, build our delta inequality out of the epsilon one. So for $$\large\rm \delta<1$$ we have:$\large\rm |x-c|<1$$\large\rm -1<x-c<1$$$\large\rm c-1<x<c+1$$ Let's call this equation (1).

6. zepdrix

Multiplying equation (1) by c, and then taking the square root gives us,$\large\rm \sqrt{c^2-c}<\sqrt{cx}<\sqrt{c^2+c}$This helps us get rid of this dependence on x here,$\large\rm \sqrt{c^2-c}<\color{green}{\sqrt{cx}<\sqrt{c^2+c}}$

7. zepdrix

Going back, taking the square root of equation (1), and then adding sqrt(c) to all sides,$\large\rm \sqrt{c-1}+\sqrt c<\sqrt{x}+\sqrt{c}<\sqrt{c+1}+\sqrt{c}$This also helps us to get rid of that pesky x!$\large\rm \sqrt{c-1}+\sqrt c<\color{royalblue}{\sqrt{x}+\sqrt{c}<\sqrt{c+1}+\sqrt{c}}$

8. zepdrix

So we have that when $$\large\rm \delta<1$$, $\large\rm \left\lvert x-c\right\rvert\quad<\quad\epsilon\color{royalblue}{\left(\sqrt x+\sqrt c\right)}\color{green}{\sqrt{cx}}\quad<\quad\epsilon\color{royalblue}{\left(\sqrt{c+1}+\sqrt{c}\right)}\color{green}{\sqrt{c^2+c}}$

9. zepdrix

$\large\rm |x-c|<\epsilon(\sqrt{c+1}+\sqrt{c})\sqrt{c^2+c}$

10. zepdrix

Let $$\large\rm \delta=\min\left\{1,~~\epsilon(\sqrt{c+1}+\sqrt{c})\sqrt{c^2+c}\right\}$$

11. zepdrix

I'm not really sure what we would say about $$\large\rm \delta>1$$ though, hmm... Hopefully I did something right in there though D: Sorry I'm kinda new at these types of problems, just trying to see if I could work through it properly. I think that proves it for $$\large\rm \delta<1$$.