Calculating the limit of

$f(x)= \dfrac{|x+1||x-1|}{x^2 -1}$

Calculating the limit as $x\Rightarrow -1^-, -1^+, 1^-, 1^+$

So mod mean you can do x+1 and 1-x cus that's how mod works.

So you can get the answers 1 and -1 only out f(x) ensuring that the original f(x) is exactly the same as 1 or -1

however i'm not sure as how to determine the limits.

You can split the range into 3 sections; $(-\infty,-1) \cup (-1,1) \cup (1, \infty)$

Then for $(-\infty, -1)$ you have $f(x)=\frac{(x+1)(x-1)}{x^2-1}$

Then for $(-1,1)$ you have ... ?

etc...

Then you can determine the limits quite simply.

how do you get $f(x)=\frac{(x+1)(x-1)}{x^2-1}$ for $(-\infty, -1)$ i'm not even sure what you do to the modulus because i dont' know what to do for -1 and 1

Your values for the domains at the end are correct.

For the one I worked out, it's quite simply the case of taking $|x+1|$ and $|x-1|$ individually, and seeing what equation without the modulus they are equivalent to on the domain $x \in (-\infty, -1)$. Clearly, you (should) know that $|x+1|=-(x+1)$ and $|x-1|=-(x+1)$, and so the product of the two must just be $|x+1||x-1|=(x+1)(x-1)$, and that's the numerator for that domain. Then just repeat similarly for the other two to deduce the answers you found.

so whatever sign the 2 defined ranges of where x can be that's what sign i stick in front?

No... do you know how the graphs of $|x+1|$ and $|x-1|$ look like? You must've sketched them at A-Level at least.

yup, they look like V and the x-1 is a V a bit more to the right

Yeah but you do realise that $|x-a|$ is the same thing as saying $x-a$ for $x\geq a$, and $-(x-a)$ for $x<a$?

Same thing applied here.

Yup ok so how the -1,1 work?

Well.. what's is the function $|x+1|$ equivalent to on this interval? What about $|x-1|$..?

0 and 0 if i stick -1 and 1 in?

OK... not quite... Got absolutely NO idea why you're sticking -1 and 1 in there... (unless I misunderstood what you've asked)

On $(-1,1)$ we have $|x+1|=x+1$ and $|x-1|=-(x-1)$

So $f(x)=\frac{-(x-1)(x+1)}{x^2-1}=-1$ on $x \in (-1,1)$ as you found.

Honestly don't know how much more basic I can go on this.

It's just this line which i don't get.
Do i need to like look at a graph of it or something?
the graph of |x+1| between -1 and 1 is oh i see?

so if i make whatevers inside || and make it 0 then i can see what the equation is?

A graph would help understand, yes.

Also going back to the definition of the modulus; you have $\displaystyle |x+1|= \begin{cases}x+1, & x \in (-1, \infty) \\ -(x+1), & x \in (-\infty,-1] \end{cases}$

Since $x \in (-1,1) \subset (-1,\infty)$, you have $|x+1|=x+1$ in this interval.

Get it?

Your values for the domains at the end are correct.

For the one I worked out, it's quite simply the case of taking $|x+1|$ and $|x-1|$ individually, and seeing what equation without the modulus they are equivalent to on the domain $x \in (-\infty, -1)$. Clearly, you (should) know that $|x+1|=-(x+1)$ and $|x-1|=-(x+1)$, and so the product of the two must just be $|x+1||x-1|=(x+1)(x-1)$, and that's the numerator for that domain. Then just repeat similarly for the other two to deduce the answers you found.

hol on a sec how come the |x+1||x-1| in the interval -infinity and -1 are both the same thing? wouldn't it be -(x-1) and -(x+1) ??

Sorry, my mistake, $|x-1|=-(x-1)$ in that interval, as you say. Their product still remains positive as stated.

https://cdn.discordapp.com/attachments/322819040059850757/379037087078416395/unknown.png

so these notes from the teacher wrong too?

Yes, that highlighted part with $|x+1|$ is incorrect. I recommend dropping them an email about it so they can amend it as required. The rest looks fine.