Question

1)test review
let f(x)=1/sqrt(x) and 0 10 years ago

Answer 1

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| \[<\delta\]

Answer 2

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.

Answer 3

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,\[\large\rm \left\lvert\sqrt x-\sqrt c\right\rvert\left\lvert\sqrt x+\sqrt c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert\]\[\large\rm \left\lvert x-c\right\rvert<\epsilon\left\lvert\sqrt x+\sqrt c\right\rvert\left\lvert\sqrt{cx}\right\rvert\]

Answer 4

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

Answer 5

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).

Answer 6

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,\[\large\rm \sqrt{c^2-c}<\color{green}{\sqrt{cx}<\sqrt{c^2+c}}\]

Answer 7

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!\[\large\rm \sqrt{c-1}+\sqrt c<\color{royalblue}{\sqrt{x}+\sqrt{c}<\sqrt{c+1}+\sqrt{c}}\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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 \(\large\rm \delta<1\).