I was wondering if I did this correctly...
No, for many reasons.
6. Calculate the indicated limit: (a)
x→−3limx2−9x2+3x
My working:
(a)
x→−3limx2−9x2+3x
(−3)2−9(−3)2+3(−3)=9−99−9=00
This last line above confirms that you
must take the limit, because you cannot evaluate.
(x−3)(x+3)x(x+3)=
This is the factorization you did back in algebra when you were working with rational functions, doing the graphing "with the hole", where you had one x-value for which the original function was not defined. But this factorization is in no way "equal" to the limit that follows. Many graders would stop grading at this "equals", with everything following being ignored as incorrect.
x→−3lim(−3)−3−3=−6−3
No. Since the expression is not, as you showed, actually defined at x = -3, you cannot take the limit by evaluating at x = -3; the function is not defined at x = -3. Instead, you have to use valid mathematical logic, such as "anything divided by itself, as long as it isn't zero, has '1' as its limiting value". This is how limits get around the "not defined at x = (whatever)" problem.
Your book (and your class notes) will contain at least one worked example of this sort. Review it for pointers:
. . . . .\(\displaystyle \begin{align} \displaystyle \lim_{x\, \rightarrow\, -3}\, \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\, &=\, \lim_{x\, \rightarrow\, -3}\, \dfrac{x\, (x\, +\, 3)}{(x\, -\, 3)\, (x\, +\, 3)}\,
\\ \\ &=\, \lim_{x\, \rightarrow\, -3}\, \left(\dfrac{x}{x\, -\, 3}\right)\, \left(\dfrac{x\, +\, 3}{x\, +\, 3}\right)\end{align}\)
...and so forth.