# Finding domain when using continuity to evaluate a limit

• I
• ChiralSuperfields
In summary: But thenIf ##x - 1...x + 1## is not in the domain of the function, then there is a value of x for which the function does not exist.In other words, if ##x - 1...x + 1## does not belong to the domain of the function, then the function does not exist at that point in the domain.This is because the function only exists for values of x that are in the domain.So, the two statements are not equivalent.The first statement implies that there is a value of x for which the function does not exist, while the second statement says that the
ChiralSuperfields
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!

Callumnc1 said:
However, I don't understand how ##x ≥ 1 -\sqrt{5}## and ##x ≥ \sqrt{5} + 1## can both be true.

ChiralSuperfields
Callumnc1 said:
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?

ChiralSuperfields
PeroK said:

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

Many thanks!

pasmith said:
If $|x| = 5$, are both $x = 5$ and $x = -5$ true simultaneously?

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

Many thanks!

Callumnc1 said:

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?

ChiralSuperfields
pasmith said:
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?

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

Many thanks!

Callumnc1 said:
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."

ChiralSuperfields
Mark44 said:
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."

I'll do some more thinking.

Many thanks!

Callumnc1 said:

I'll do some more thinking.

Many thanks!
What is it that you are thinking about ?

ChiralSuperfields and PeroK
Callumnc1 said:
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)##.

ChiralSuperfields and PeroK
SammyS said:
What is it that you are thinking about ?

Many thanks!

Mark44 said:
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)##.

I will review that notation more.

Many thanks!

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