What is the domain of this function (a limit probably)?

[Koalanet21]

(2x^2+x-3)/(x^2+x-2)

So I can see what is the method and learn it for other similar functions

 

[R.M.]

To find the domain, start with this question: where is the function undefined?

 

[Deleted member 4993]

To find the domain, start with this question: where is the function undefined?

To do that - I would factorize the numerator and the denominator.

 

[Deleted member 4993]

(2x^2+x-3)/(x^2+x-2)

So I can see what is the method and learn it for other similar functions

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[JeffM]

And a domain is never a limit.

 

[Koalanet21]

I calculated the "conditions of existence"* and found x couldn't be equal to 1 in both equations. So maybe lim->1. If x=1, then both functions would be equal to zero. So I did b²=4ac in both and simplified by (x-1), replaced all x by 1 and got 5/3

But then I have no idea how to get the domain of the function, I don't even know where to start.
Should I use the roots of the 2 functions (denominator and numerator)? But which one of them? -3 and 1 or -2 and 1 ?

Condition d'existence — Wikipédia

fr.wikipedia.org

*I actually don't know how to call that in English, and couldn't find any reliable translation on the Internet. Even the wikipedia page doesn't not have any interlanguage link. Are the conditions of existence peculiar to French maths? Or is it just not named at all in English?

 

[lex]

I think you have found that [MATH]\frac{(2x^2+x-3)}{(x^2+x-2)}=\frac{(2x+3)(x-1)}{(x+2)(x-1)}[/MATH]The 'conditions for existence' (to use your phrase) for a fraction like this is that the bottom line is not 0.
That gives you your (largest possible) domain. (All x values, except those where the bottom line will become 0).

 

[HallsofIvy]

The fraction \(\displaystyle \frac{(2x+3)(x- 1)}{(x+ 2)(x-1)}\) has denominator \(\displaystyle (x+ 2)(x- 1)\) which is 0 at x=-2 and x= 1, The domain is "all real numbers except 1 and -2".

Don't be fooled by the "x- 1" in the numerator.
\(\displaystyle \frac{(2x+3)(x- 1)}{(x+ 2)(x-1)}= \frac{2x+ 3}{x+ 2}\)
only if \(\displaystyle x\ne 1\). If x= 1 the right side is 5/3 while the left side does not exist.
The right side has domain "all real numbers except x= -2" while the left side has domain "all real numbers except x= -2 and x= 1".

 

[lookagain]

To do that - I would factorize the numerator and the denominator.

I do not see factorizing the numerator giving more information toward the domain. It can help
to determine where there is a vertical asymptote versus a "hole" in the graph.