I Finding domain when using continuity to evaluate a limit

[member 731016]

For this problem,

The solution is,

However, when I tried finding the domain myself:

${ x | x - 1 ≥ \sqrt{5}}$ (Sorry, for some reason the brackets are not here)
${ x | x - 1 ≥ -\sqrt{5}}$ and ${ x | x - 1 ≥ \sqrt{5}}$
${x | x ≥ 1 -\sqrt{5} }$ and ${ x | x ≥ \sqrt{5} + 1}$

However, I don't understand how $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$ can both be true. Because of that, I also don't understand how they got the domain.

Many thanks!

 

[PeroK]

However, I don't understand how $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$ can both be true.

You need to know your unions from your intersections.

 

[pasmith]

However, I don't understand how $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$ can both be true. Because of that, I also don't understand how they got the domain.

Many thanks!

If |x| = 5, are both x = 5 and x = -5 true simultaneously?

 

[member 731016]

You need to know your unions from your intersections.

Thank you for your reply @PeroK!

I kind of remember, but it was from over a year ago now.

Many thanks!

 

[member 731016]

If |x| = 5, are both x = 5 and x = -5 true simultaneously?

Thank you for your reply @pasmith!

Yes I believe so (if we sub the values of x into the absolute value function).

Many thanks!

 

[pasmith]

Thank you for your reply @pasmith!

Yes I believe so (if we sub the values of x into the absolute value function).

Many thanks!

No, exactly one of those is true; x cannot satisfy both x = 5 and x = -5 simultaneously, because -5 \neq 5. But |x| = 5 will be true if either of those statements is true. Thus \{x : |x| = 5\} = \{-5\} \cup \{5\} = \{-5,5\}. How would you apply that logic to your problem?

 

[member 731016]

No, exactly one of those is true; x cannot satisfy both x = 5 and x = -5 simultaneously, because -5 \neq 5. But |x| = 5 will be true if either of those statements is true. Thus \{x : |x| = 5\} = \{-5\} \cup \{5\} = \{-5,5\}. How would you apply that logic to your problem?

Thank you for your reply @pasmith !

I will do some hard thinking and get back to you.

Many thanks!

 

[Mark44]

For this problem,
View attachment 323369
View attachment 323368
The solution is,
View attachment 323370
However, when I tried finding the domain myself:

$\{ x | x - 1 ≥ \sqrt{5}\}$ (Sorry, for some reason the brackets are not here) Now fixed..
$\{ x | x - 1 ≥ -\sqrt{5}\}$ and $\{ x | x - 1 ≥ \sqrt{5}\}$
$\{x | x ≥ 1 -\sqrt{5} \}$ and $\{ x | x ≥ \sqrt{5} + 1\}$

However, I don't understand how $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$ can both be true. Because of that, I also don't understand how they got the domain.

1. The reason the braces disappeared is that they are special characters in TeX that are used for multiple purposes (e.g., exponents, fractions, subscripts, etc.). If you need to display a left or right brace, precede it with a slash. That's what I did for the brace pairs in what you wrote above.
2. As already mentioned, the author did not write $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$. The symbol they used was $\cup$ or union, which corresponds to "or" not "and."

 

[member 731016]

1. The reason the braces disappeared is that they are special characters in TeX that are used for multiple purposes (e.g., exponents, fractions, subscripts, etc.). If you need to display a left or right brace, precede it with a slash. That's what I did for the brace pairs in what you wrote above.
2. As already mentioned, the author did not write $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$. The symbol they used was $\cup$ or union, which corresponds to "or" not "and."

Aah, ok. Thank you for your reply @Mark44!

I'll do some more thinking.

Many thanks!

 

[SammyS]

Aah, ok. Thank you for your reply @Mark44!

I'll do some more thinking.

Many thanks!

What is it that you are thinking about ?

 

[Mark44]

However, when I tried finding the domain myself:
${ x | x - 1 ≥ \sqrt{5}}$ (Sorry, for some reason the brackets are not here)
${ x | x - 1 ≥ -\sqrt{5}}$ and ${ x | x - 1 ≥ \sqrt{5}}$
${x | x ≥ 1 -\sqrt{5} }$ and ${ x | x ≥ \sqrt{5} + 1}$

However, I don't understand how $x ≥ 1 -\sqrt{5}$ and $x ≥ \sqrt{5} + 1$ can both be true.

You have another mistake that I didn't notice while I was focused on your difficulties with LaTeX and confusion about the union vs. intersection of two sets.

Starting with $|x - 1| \ge \sqrt 5$,
This means that $x - 1 \le -\sqrt 5$ OR $x - 1 \ge \sqrt 5$
$\Rightarrow x \le 1 - \sqrt 5$ OR $x \ge 1 + \sqrt 5$.

The second line of your work above that I quoted is incorrect because you have not correctly rewritten the inequality with an absolute value to get rid of the absolute value. Again, this is stuff that is usually presented in precalculus classes. Until you get a better grip on these basics, you are going to continue to have problems with more advanced topics.

The last pair of inequalities that I wrote can be written in set-builder notation like so:
$\{x | (-\infty < x \le 1 - \sqrt 5) \cup (1 + \sqrt 5 \le x < \infty)\}$

or in interval notation like this:
$x \in (-\infty, 1 - \sqrt 5] \cup [1 + \sqrt 5, \infty)$.

 

[member 731016]

What is it that you are thinking about ?

Thank you for your reply @SammyS!

I am just thinking more about it before I ask questions.

Many thanks!

 

[member 731016]

You have another mistake that I didn't notice while I was focused on your difficulties with LaTeX and confusion about the union vs. intersection of two sets.

Starting with $|x - 1| \ge \sqrt 5$,
This means that $x - 1 \le -\sqrt 5$ OR $x - 1 \ge \sqrt 5$
$\Rightarrow x \le 1 - \sqrt 5$ OR $x \ge 1 + \sqrt 5$.

The second line of your work above that I quoted is incorrect because you have not correctly rewritten the inequality with an absolute value to get rid of the absolute value. Again, this is stuff that is usually presented in precalculus classes. Until you get a better grip on these basics, you are going to continue to have problems with more advanced topics.

The last pair of inequalities that I wrote can be written in set-builder notation like so:
$\{x | (-\infty < x \le 1 - \sqrt 5) \cup (1 + \sqrt 5 \le x < \infty)\}$

or in interval notation like this:
$x \in (-\infty, 1 - \sqrt 5] \cup [1 + \sqrt 5, \infty)$.

Thank your reply @Mark44!

I will review that notation more.

Many thanks!