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Define $f(x)=\sqrt{x}$ for all $x\geq 0$. Verify the $\epsilon,\delta$ criterion for continuity at x=4 and at x=100. Hint: first show that for $x\geq0$, $x_0\gt0$, $|\sqrt{x}-\sqrt{x_0}|\leq|x-x_0|/\sqrt{x_0}$.

For my proof (just for $x=4$), I used that $\forall x \geq0, \sqrt{x} \leq x$. So, I let $\delta = \epsilon$, and then I let $x \in \mathbb{R}$ be such that $|x-4| \lt \delta$. Then, $|\sqrt{x}-\sqrt{4}|\leq|x-4|\lt \delta = \epsilon$. Is this a valid statement? I feel a little "iffy" about the validity of going from my $|\sqrt{x}-\sqrt{4}|$ to $|x-4|$.

I am also confused about the hint that the problem gave. Why would I need to prove that first? (In class, we just do the problems applying the $x_0$ value immediately.)

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  • $\begingroup$ Note: $\forall x \ge 0; \sqrt{x} \le x$ is not true. $\forall x \ge 1 \sqrt x \le x$ but $\forall x: 0 \le x \le 1; \sqrt{x} \ge x$. $\endgroup$ – fleablood Sep 14 '18 at 15:55
  • $\begingroup$ Why would $x \le w$ and $y\le z$ imply that $|x - y| \le |w-z|$? That is not a valid statement at all. $\endgroup$ – fleablood Sep 14 '18 at 15:57
  • $\begingroup$ Well, if you know $|\sqrt{x} - \sqrt 4|\le \frac {|x-4|}{\sqrt{x}}$ you can use it immediately (I guess) but you have to show that first. $\endgroup$ – fleablood Sep 14 '18 at 16:03
  • $\begingroup$ "Is this a valid statement?" 1) It's only true for $x,4 \ge 1$ and 2) Although it is a true statement for $x,4 \ge 1$ it is not a valid statement if no reason is given. ANd $\sqrt{x} \le x$ and $\sqrt 4 \le 4$ is not a valid reason. (Obviously $5< 24$ and $7 < 25$ but $|7-5| > |25-24|$.) $\endgroup$ – fleablood Sep 14 '18 at 16:31
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Guide:

  • As mentioned in the comment, $\forall x \ge 0, \sqrt{x} \le x$ is not true.

  • $|\sqrt{\frac14} - \sqrt{\frac1{16}} | =\frac12-\frac14=\frac14=\frac{4}{16} $ but $\frac14 - \frac1{16}=\frac{3}{16}$

  • If you have the hint, then your proof can be as follows.

For example, let $x_0=4$, then

If $0< |x-4|< \delta$, then $|\sqrt{x}-\sqrt{4}| \le \frac{|x-4|}{\sqrt4}\le \frac{\delta}2$ and you can choose your $\delta$ in terms of your $\epsilon$ easily.

  • Useful tools to prove the hint: $a^2-b^2 = (a-b)(a+b)$ and also if $a>0, b \ge 0$, then $\frac{1}{a+b} \le \frac1a$ .
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  • $\begingroup$ But for $x,y \ge 1$ then is true that $|\sqrt{x} - \sqrt{y}| \le |x - y|$ So the argument would be valid if the OP had claimed $\forall x\ge 1; \sqrt x \le x$. But $|\sqrt{x} - \sqrt{y}| \le |x - y|$ must be proven.... or at least explained. $\endgroup$ – fleablood Sep 14 '18 at 16:17
  • $\begingroup$ yup, I agree. also it can be a consequence of the hint $\endgroup$ – Siong Thye Goh Sep 14 '18 at 16:19
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    $\begingroup$ Yes. If $x,y\ge 1$ then $|\sqrt x - \sqrt y| \le \frac {|x-y|}{\sqrt y} \le |x-y|$ so the OPs original idea would be much less efficient. But it would worked (had s/he successfully prove his/her claim).... however for $0< y < 1$ something else would have to be done as $|\sqrt {x} - \sqrt{y}| > |x - y|$. In which case the hint is good as $\frac {|x-y|}{\sqrt y} \le |x-y|$ would still be true for all $x \ge 0$.... I'm a bit surprised the book didn't through in an "at $x = \frac 19$" just to toss a curve ball. $\endgroup$ – fleablood Sep 14 '18 at 16:25

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