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.
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 \displaystyle \lim_{x\rightarrow 2} \frac{x^2-4}{x-2} $
$\displaystyle \displaytype\lim_{x\to a} f(x)$ exists if and only if $\displaystyle \displaytype\lim_{x\to a^+} f(x)$ and $\displaystyle \displaytype\lim_{x\to a^-} f(x)$ both exist and are the same.
Yes, $\displaystyle \displaytype\lim_{x\to 2^-} f(x)= \displaytype\lim_{x\to 2^-}\frac{x^2- 4}{x- 2}= 4$. That is "$\displaystyle x\to 2^-$, 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": $\displaystyle \displaytype\lim_{x\to 2^+} f(x)= \displaytype\lim_{x\to 2^+} ax^2- bx+ 3= 4x- 2b+ 3$ must equal 4. That gives you one equation.
Do the same with $\displaystyle \displaytype\lim_{x\to 3^-} ax^2- bx+ 3$ and $\displaystyle \displaytype\lim_{x\to 3^+} 3x- a+ b$ to get a second equation.