Why am I not getting an infinite limit? I am trying to solve $\displaystyle\lim_{x\to3^-}\frac{3}{x - 3} - \frac{3}{x} - 9$. Here's what I have tried.
$$
\lim_{x\to3^-}\frac{3}{x - 3} - \frac{3}{x} - 9 \\
\lim_{x\to3^-}\frac{3x}{x(x - 3)} - \frac{3(x-3)}{x(x-3)} - \frac{9x(x-3)}{x(x-3)} \\
\lim_{x\to3^-}\frac{3x - 3x - 9x^2 + 27x}{x(x - 3)} \\
\lim_{x\to3^-}\frac{- 9x^2 + 27x}{x^2 - 3x)} \\
$$
By L' Hopital's Rule,
$$
\lim_{x\to3^-}\frac{-18x + 27}{2x - 3} \\
\lim_{x\to3^-}\frac{-18}{2} \\
-9 \\
$$
However, Wolfram Alpha claims that the limit is infinity:
http://www.wolframalpha.com/input/?i=lim+x+-%3E+3+%5Cfrac%7B3%7D%7Bx+-+3%7D+-+%5Cfrac%7B3%7D%7Bx%7D+-+9
Why?
 A: In your second step, $3(x-3)=3x-9$, not $3x$. When you restore the missing term, you’ll find that l’Hospital’s rule no longer applies.
A: You made an algebra mistake; instead of
$$\lim_{x\to3^-}\frac{3x - 3x - 9x^2 + 27x}{x(x - 3)}$$
It is
$$\lim_{x\to3^-}\frac{3x - 3x + 9 - 9x^2 + 27x}{x(x - 3)}$$
So you just forgot a term.
A: You missed a term $+9$ on the numerator of your fraction. This matters because then the limit is not of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, which is a hypothesis of l'Hôpital's rule.
A: The numerator and denominator of the fraction will both be divisible by $x-3$ if the limit exists.  You'll have
$$
\lim_{x\to3} \frac{(x-3)(\cdots\cdots\cdots)}{(x-3)(\cdots\cdots)}.
$$
Then cancel the common factor.  After that, you can plug in $3$ for $x$.
A: Why not use WolframAlpha to help you find where the problem occurred?  Your third line already evaluates to -9, so you know the error is not in the rest of your computation.
In fact the plot shows that your third line is identically equal to -9, which hints that you did something to incorrectly cancel out the $3/(x-3)$ and the $3/x$.  I'm being deliberately vague here to highlight the fact that this basic level of information is in your grasp.
