Issues Calculating Limits Using Derivatives I'm pretty bad at math, and Im doing some practice questions for school, I was wondering if someone could help me with this question, I cant get the right answer no matter what I try. Here's the question:
$$
f(x) = \left\{\begin{array}{ll}
-\frac{3}{x+3},&\text{if } x \lt -3;\\
2x+9, &\text{if }x \gt -3.
\end{array}\right.$$
Calculate the following limits:
a) $\displaystyle \lim_{x\to -3^-} f(x)$
b) $\displaystyle \lim_{x\to -3} f(x)$ 
c) $\displaystyle \lim_{x\to -3^+} f(x)$ 
I've gotten (These are right)
B) DNE
C) 3
But I cant figure out A do I need the
$$\frac{f(x + h) - f(x)}{h}$$ 
or 
$$\frac{f(x)-f(a)}{x-a}$$ for that one? how would I work it out?
Here's my work if you want to look at what i did wrong:

 A: In what you write, you are not being asked to do a derivative, you are just being asked to do a limit. So you do not need to use either the Fermat or the Difference quotient limits. You just need to do the limit of the function.
As $x\to -3^-$, $f(x)$ will be evaluated using the formula $\displaystyle-\frac{3}{x+3}$. So
$$\lim_{x\to -3^-}f(x) = \lim_{x\to -3^-}-\frac{3}{x+3}.$$
When $x\to -3^+$, $f(x)$ is evaluated using the formula $2x+9$, so you would do
$$\lim_{x\to -3^+}f(x) = \lim_{x\to -3^+}(2x+9).$$
These limits should be done directly; neither of the derivative limits comes into play at all. 
If you are being asked for the limits of the derivative of $f(x)$, i.e.,
$\displaystyle\lim_{x\to -3^-}f'(x)$, $\displaystyle\lim_{x\to -3^+}f'(x)$, $\displaystyle \lim_{x\to -3}f'(x)$, then first you should figure out the derivative for each of the two "parts" of $f(x)$ using whichever quotient you want, then take the limit.
For example, if $a\gt -3$, then
$$
f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x\to a}\frac{(2x+9)-(2a+9)}{x-a};$$
or, just as valid,
$$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0}\frac{2(a+h)+9 - (2a+9)}{h}.$$
Then, whatever the answer is, you use this formula for $\displaystyle \lim_{x\to -3^+}f'(x)$.
Similarly with the other limit. 
A: When stuck with limits, sometimes drawing a graph could help.
as x approaches -3 from the negative side, the function -(3/(x+3)) approaches very large values of y. As x approaches -3 from the right, the function approaches 2*-3+9=3. 
As a result the function has no limit at x=-3 and is not continuous at this point.
In an exam, you probably can't relay on the plot, but the picture together with algebraic derivation sometimes help.
Wolfram provide a free on-line plotter. Also, you may want to check this: GraphPlotter
