Question

Ruddy Epsilon-delta proofs... I need a hint on how to bound \(|\sqrt{x}+\sqrt{c}| \) s.t. \(c\ge 0\)

Answer 1

\(|\sqrt{x}+\sqrt{c}| \) s.t. \(c\ge 0\)

Answer 2

@Hero Could you take a look for me please?

Answer 3

Do you want my prior work?

Answer 4

What's the actual limit?

Answer 5

the problem is:\[\lim_{x\rightarrow c}\sqrt{x}~~~~for ~x\ge0\]\[L=\sqrt{c}\]

Answer 6

I have it modified to here : \[|x-c|<\frac{\epsilon}{|\sqrt{x}+\sqrt{c}|}\]

Answer 7

Usually, I'd say |x-a|<1, but since it is a c, I don't know how to extrapolate a general form

Answer 8

Small mistake, you should have \[|\sqrt x-\sqrt c|\cdot\frac{|\sqrt x+\sqrt c|}{|\sqrt x+\sqrt c|}=|x-c|\frac{1}{|\sqrt x+\sqrt c|}<\epsilon~~\iff~~|x-c|<|\sqrt x+\sqrt c|~\epsilon\]
Anyway, we still need to find a bound for that \(|\sqrt x+\sqrt c|\) expression...

We know that \(\sqrt x\) is defined for \(x\ge0\), which means \(|\sqrt x+\sqrt c|\ge\sqrt c\). So,
\[|x-c|\frac{1}{|\sqrt x+\sqrt c|}\le|x-c|\frac{1}{\sqrt c}<\epsilon~~\iff~~|x-c|<\sqrt c~\epsilon\]

Answer 9

wait... that is what I have?

Answer 10

oh goodness nvm on the mistake... I'm a nitwit

Answer 11

so why can we just drop the √x?

Answer 12

because we want the max?

Answer 13

\(\sqrt{x}\) sorry

Answer 14

At the very least, when \(x=0\), you have \(\sqrt{x}+\sqrt{c}=0+\sqrt c=\sqrt c\), which means
\[\sqrt x+\sqrt c=\sqrt c\]
For any other value of \(x\) in the domain of the square root function, you will necessarily have
\[\sqrt x+\sqrt c>\sqrt c\]
This means \(\sqrt x+\sqrt c\) is bounded below by \(\sqrt c\), i.e. \(\dfrac{1}{\sqrt x+\sqrt c}\) is bounded above by \(\dfrac{1}{\sqrt c}\).

Answer 15

ah ok I get it now, thanks!

Answer 16

You're welcome!