Find a $\delta > 0$ such that if $|x-3|<\delta $, then $|\sqrt{2x+3}-3|<0.01$.

Question

Could someone correct me if I made a mistake? The textbook's answer is $0.02995$.

Find a $\delta > 0$ such that if $|x-3|<\delta $, then $|\sqrt{2x+3}-3|<0.01$.

$-0.01\leq \sqrt{2x+3}-3\leq 0.01$

$2.99\leq \sqrt{2x+3}\leq 3.01$

$8.94\leq 2x+3\leq 9.06$

$5.94\leq 2x\leq 6.06$

$2.97\leq x\leq 3.03$

$-0.03\leq x-3\leq 0.03$ (in the textbook, it was written $3-0.02995\leq x-3\leq 0.03$ and I am not sure why)

So would $\delta = 0.03$ be okay instead of $\delta = 0.02995$?

Answer 1

With your $0.03$, for $x=3-.03=2.97$, you get $\sqrt {2x+3}-3 =-.010016722$ which is not acceptable. The smaller $\delta $ is always the safe way to go.

Answer 2

The negative epsilon will yield the smaller $\delta$

$\sqrt{2x+3}-3 = -.01$

$\sqrt{2x+3} = 2.99$

$2x+3 = 8.9401$

$2x = 5.9401$

$x = 2.97005$

$\delta < .02995$

Answer 3

Let's find out:

$|x- 3| < .03$

$-.03 < x -3 < .03$

$2.97 < x < 3.03$

$5.94 < 2x < 6.03$

$8.94 < 2x + 3 < 9.03$

$2.9899832775452106197018615265259 < \sqrt{2x + 3} < 3.0049958402633438587273801370837$

$-0.01001672245478938029813847347409 < \sqrt{2x+3} - 3 < .0049958402633438587273801370837\not \implies$

$|\sqrt{2x+3}-3|< 0.01$ as $2.9899832775452106197018615265259 < \sqrt{2x + 3} < 2.99$ would cause that to fail.

Your error was rounding $2.99^2 = 8.9401$ down. We must have $x$ in this range. So although it okay to roll the range narrower (as you did when you rolled $9.0601$ down) you can't roll it wider.

So $5.9401 \le 2x \le 6.0601$

$2.97005 \le x \le 3.03005$

$-0.02995 \le x \le 0.03005$. And $\delta \le \min(|-0.02995|, |0.03005|)$ is required.

Note. there is utterly no need to find the largest possible $\delta$.

I personlly would have down.

$8.9401 \le 2x + 3 \le 9.0601 \Leftarrow$

$8.95 < 2x + 3 < 9.05$

$5.95 < 2x < 6.05$

$-.05\le |2x - 6| < .05$

$|2x - 6|<0.05$

$|x - 3| < 0.025$ and simply have set $\delta = .025$. That's not the largest possible $\delta$. But it is a a delta that works.