Limit trouble! Help please

[johnnyICON]

I have this question from a past assignment that I never get right, it says the limit is \frac{1}{2}.
But what I keep getting is undefined.

Here is the problem: \lim_{x \to 1} \frac{\x^2 - 4x + 3}{\-x^2 + 2x + 3}.

I broke it down using limit laws, then substituted in 1 for the x's, and ended up with \frac{0}{0}. What am I doing wrong here? :confused:

[Math Is Hard]

Did you try doing any factoring of numerator and denominator first?

[johnnyICON]

Did that as well, I got \frac{(x-1)(x+4)}{(-x+1)(x+3)}. And when you substitute 1 in, its still \frac{0}{0}.

[Embermage]

I think you factored the top wrong... I get (x-3)(x-1)... that should help you out, because I believe when you cancel (x-3) with the (x+3) on the bottom you get -1. I think.

[Math Is Hard]

\lim_{x \to 1} \frac{x^2 - 4x + 3}{\-x^2 + 2x + 3}

I am having a little bit of trouble reading your markup. Is this what we're working on?

[Math Is Hard]

or is that squared x term on the bottom negative?

\lim_{x \to 1} \frac{x^2 - 4x + 3}{\ (-)x^2 + 2x + 3}

I am having trouble getting that negative sign to show up.

[johnnyICON]

or is that squared x term on the bottom negative?

\lim_{x \to 1} \frac{x^2 - 4x + 3}{\ (-)x^2 + 2x + 3}

I am having trouble getting that negative sign to show up.


This is the one. By the way, I've been meaning to ask how to make things look bigger?

I use the itex tags, your using tex tags? Anyway, get back to me on that one later

[johnnyICON]

OH MY !!! :eek:
I'm an idiot... I factored wrong... as you may have noticed, my factored numerator is magically (x-1)(x+4)... :( It should be (x-1)(x-3) right?

[johnnyICON]

And I think I'm wrong by including the sign...

Should it be \frac{(x-1)(x-3)}{-(x-1)(x-2)} rather than \frac{(x-1)(x-3)}{(-x-1)(x+2)} ...?

[plover]

For the expression you gave -(x+1)(x-3) would be the correct factoring for the denominator.

However:
\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 + 2x + 3)}\ =\ \frac{0}{4}\ =\ 0
Are you sure you've got the right expression? The problem is simple unless 1 is actually a root of both the numerator and the denominator.
\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \frac{1}{2}
This one works out to the answer you indicated. You can either factor or use l'Hôpital's rule.

[plover]

Note on tex:

The reason the minus sign is disappearing is due to putting a backslash in front of it.
Incorrect: \-x^2+2x+3

Correct: -x^2+2x+3

[johnnyICON]

I know. I keep on double checking to see whether it wasn't just a smudge of ink or pencil mark, but yes, the sign is included...

For your first equation, did you mean numerator instead of denominator?

And how did you get 1/2 for the second one?

Here's what I've been getting
r(x) =\ \frac{(x-1)(x-3)}{-(x-1)(x-2)} =\ \frac{x-3}{-(x-2)} =\ \frac{1 - 3}{-(1-2)} =\ \frac{-2}{1}...??

It seems like I get farther and farther way everytime I try something else...

[plover]

If your original expression is identical to the one in your book then this is a typo in the book. The limit of the original expression is 0 not 1/2.

Also:
Yes, I do mean the denominator. If a minus sign in front of the squared term makes the factoring difficult, put the expression in a form where you don't need to worry about it:
-x2 - 2x + 3 = -(x2 + 2x - 3)
The factors you give are incorrect:
(x-1)(x-2) = x2 - 3x + 2

[johnnyICON]

DFDKJSDFJFLdfkdfoiaanog Wow, I always manage to read something wrong... ok but wait, doesnt (x+1)(x-3) =\ x^2 - 2x - 3?

[plover]

Yes,
(x + 1)(x - 3) = x2 - 2x - 3
so these are not the correct factors. Keep in mind that it is necessary for 1 to be a root of the polynomial—with your factors, the roots would be -1 and 3.

[johnnyICON]

Alright...

:D

Can you show me how you did this:

\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \frac{1}{2}


That's with the denominator being (-x+1)(x+3) right?

[plover]

Yes, that is the denominator. It's generally best though to pull the negative sign outside of the factors: -(x-1)(x+3)

So, you've got the expression factored like this:

\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \lim_{x \to 1} \frac{(x-1)(x-3)}{-(x-1)(x+3)}

Do you see how to proceed from there?

[johnnyICON]

ahh... !!!!! I'm an idiot. THANKS :D

[johnnyICON]

I think where my confusion lies is with that damn negative sign.

It should always be (-x)^2 -2x + 3 unless stated as -(x^2 -2x + 3)...

Trying to make a rule for myself.. do you agree?

[plover]

The standard rule is that exponents have precedence over multiplication and division, and the latter have precedence over addition and subtraction.

So -x^2+2 without any other notations is always -(x2)+2

This is used universally, so adopting any other convention will just lead to confusion.

Also, to use the convention you give (i.e. addition and subtraction have precedence over exponents) consistently you would end up with:
-x^2+2 = (-x)(2+2)

[johnnyICON]

Thanks. =D