When tackling this problem, parting from the assumption that $\lvert \sqrt{x-5} - 2\rvert \lt \epsilon$, I arrived through arithmetical manipulation at $$\frac{\lvert x-9\rvert}{\lvert \sqrt{x-5} + 2\rvert} \lt \epsilon,$$ which has the $\lvert x-9\rvert$ needed to prove that for a certain $\delta$, $0<|x-9|<\delta \Rightarrow |\sqrt{x-5} + 2|<\epsilon$. However, I don't know how to eliminate the $x$ in ${\lvert \sqrt{x-5} + 2\rvert}$, such that I can define $\delta$ solely based on $\epsilon$. I would appreciate any help.
Asked
Active
Viewed 136 times
1
-
3Should the denominator be $|\sqrt{x-5}+2|$ instead? – player3236 Oct 22 '20 at 08:07
-
@player3236 Yes, sorry about that, it was a typo. – Gabriel Nohama Oct 22 '20 at 08:13
-
4In that case, do you see that the denominator must be larger than $2$? This will help you define your $\delta$. – player3236 Oct 22 '20 at 08:14
-
Have a look at the examples on this page. Once you find a suitable value of $\delta$ in terms of $\epsilon$, you need to compile what you have written into a proper proof. – Toby Mak Oct 22 '20 at 08:17
-
@player3236 Oh I see, but that means that $\frac{|x-9|}{|\sqrt{x-5}+2|} < \frac{|x-9|}{2}$ right? But can I say that $\frac{|x-9|}{2} < \epsilon$? – Gabriel Nohama Oct 22 '20 at 08:21
-
If you choose an appropriate $\delta$. We have $\frac{|x-9|}2<\frac \delta 2$, and it can be less than $\epsilon$. Be careful though, you don't want $x-5$ to be able to be negative. – player3236 Oct 22 '20 at 08:23
-
2@player3236 Oh ok I think I understand it. Thank you! – Gabriel Nohama Oct 22 '20 at 08:27
2 Answers
0
Hints: For this particular problem , for convenient first we'll calculate the value of $\delta$ ,
We have $$|\sqrt{x-5}-2|<\epsilon$$
$$-\epsilon<\sqrt{x-5}-2<\epsilon$$
$$(2-\epsilon)^2-4<x-5<(2+\epsilon)^2-4$$.
Now choose $\delta=\text{min}\{(2-\epsilon)^2-4,(2+\epsilon)^2-4$}.
Unknown
- 3,073
-
1Where did the +4 come from on both sides of the final inequality? – Prasun Biswas Oct 22 '20 at 08:26
-
-
-
1Oh I see, but shouldn't $(2-\epsilon)^2 - 4$ and $(2+\epsilon)^2 - 4$ be absolute values? – Gabriel Nohama Oct 22 '20 at 08:31
-
-
1@player3236 but since $\epsilon$ is positive and arbitrarily small, isn't it possible that $(2 - \epsilon)^2 < 4$, therefore making $(2 - \epsilon)^2 - 4$ negative? – Gabriel Nohama Oct 22 '20 at 08:45
-
-
-
The third inequality should have $x-9$ in the middle. $\delta$ should be chosen as $\min{4-(2-\epsilon)^2,(2+\epsilon)^2-4}$ – Shubham Johri Oct 22 '20 at 08:54
-
Yes, I had just noticed that $x - 5$, this makes more sense. Thanks! – Gabriel Nohama Oct 22 '20 at 09:05
0
Assuming $|x-9|<\delta$ we have $$ \left| \sqrt{x-5} - 2 \right| = \left| \frac{(\sqrt{x-5} - 2)(\sqrt{x-5} + 2)}{\sqrt{x-5} + 2} \right| = \left| \frac{x-9}{\sqrt{x-5} + 2} \right| < \frac{\delta}{\sqrt{x-5} + 2} . $$ Now note that $\frac{1}{\sqrt{x-5} + 2} < \frac{1}{2}$ as long as $x-5>0.$ The latter can be achieved by making sure that $|x-9|<4.$ Therefore we will require $\delta<4.$ Then we can just take $\delta\leq 2\epsilon$ and we can continue: $$ \cdots < \frac{\delta}{2} \leq \epsilon. $$
Thus, given $\epsilon>0$, take $\delta=\min(2\epsilon,4).$
md2perpe
- 26,770