Continous Question

**stiitches** · September 30th 2010, 06:55 PM

I need help on this problem:

Also please explain how to do it. I started out by graphing the first function but not sure how to do the others/what its asking for.

**Gusbob** · September 30th 2010, 07:05 PM

This looks like an assignment of some sort so I won't give direct answers.

By definition of conitnuity, limit x-> a f(x) = f(a)

The key here is to find a,b such that the limit from the left of 2 and the limit from the right of 2 correseponds to f(2).
Basically you'll need to find this:

$\displaystyle \lim_{x\rightarrow 2} \frac{x^2-4}{x-2}$

**stiitches** · September 30th 2010, 07:13 PM

Alright, im still confused...

the limit of that is 4. I dont get how that relates with a and b

**HallsofIvy** · October 1st 2010, 02:30 AM

$\displaytype\lim_{x\to a} f(x)$ exists if and only if $\displaytype\lim_{x\to a^+} f(x)$ and $\displaytype\lim_{x\to a^-} f(x)$ both exist and are the same.

Yes, $\displaytype\lim_{x\to 2^-} f(x)= \displaytype\lim_{x\to 2^-}\frac{x^2- 4}{x- 2}= 4$ . That is " $x\to 2^-$ , the limit from below, because that fraction form is true only for x< 2. In order that the limit at 2 (from both directions) exist, we must get the same limit "from above": $\displaytype\lim_{x\to 2^+} f(x)= \displaytype\lim_{x\to 2^+} ax^2- bx+ 3= 4x- 2b+ 3$ must equal 4. That gives you one equation.

Do the same with $\displaytype\lim_{x\to 3^-} ax^2- bx+ 3$ and $\displaytype\lim_{x\to 3^+} 3x- a+ b$ to get a second equation.

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