<s>[INST] How do I calculate the solution to this limit algebraically? I was given a limit problem not dissimilar to this one:
Compute the limit of $\lim_{x \to -4} \frac{2x + 7}{3x^2 + 2x - 40}$
Since the numerator and the denominator have no common factors and the denominator approaches zero, the limit should approach $- \infty$, $+ \infty$, or it does not exist. For a limit like $\lim_{x \to 2} \frac{2}{2 - x}$, I can take the left and right-hand limits as follows:
$\lim_{x \to 2^{-}} \frac{2}{2 - x} = \frac{2}{2 - 2^{-}} = \frac{2}{0} = \infty$
$\lim_{x \to 2^{+}} \frac{2}{2 - x} = \frac{2}{2 - 2^{+}} = \frac{2}{-0} = - \infty$
$\lim_{x \to 2} \frac{2}{2 - x} \text{ does not exist}$
I attempted to do the same for this limit:
$\lim_{x \to -4^{-}} \frac{2x + 7}{3x^2 + 2x - 40} = \frac{-8}{48^{+} - (8^{+}) - 40}$
$\lim_{x \to -4^{+}} \frac{2x + 7}{3x^2 + 2x - 40} = \frac{-8}{48^{-} - (8^{-}) - 40}$
But I can't seem to go any farther. Looking at the graph of the equation tells you the limit does not exist, and I know that you can try substituting nearby values to -4 from both sides as well, but is there a way to compute this limit algebraically?
 [/INST] Hint: I might be helpful to consider one of the expressions of the right-hand side below.
\begin{align*}
\lim_{x\to-4}\frac{2x+7}{3x^2+2x-40}&=\lim_{x\to -4}\frac{2x+7}{(x+4)(3x-10)}\\
&=\lim_{x\to-4}\left(\frac{1}{22}\,\frac{1}{x+4}+\frac{41}{22}\,\frac{1}{3x-10}\right)
\end{align*}
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