calculus - solving limit with complex polynomial I'm try to solve $\lim \limits_{x \to -2} \dfrac{x^3 - x^2 - x + 10}{x^2 + 3x + 2}$.
I have simplified it down to $\lim \limits_{x \to -2} \dfrac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)}$.
The solution is $-15$.
However, I can't wrap my mind around it. $(x^2 - 3x + 5)$ has only complex roots because $\Delta < 0$, so how am I supposed to reach the real number $-15$ as the solution?
 A: The roots of $x^2-3x+5$ are irrelevant here. You can simplify and then simply evaluate at $x=-2$ :
\begin{align}
\lim \limits_{x \to -2} \frac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)} &= \lim \limits_{x \to -2} \frac{x^2 - 3x + 5}{x + 1} \\
&= \frac{(-2)^2-3(-2)+5}{(-2)+1} \\
&= -15
\end{align}
A: "Solve" is the wrong word here. You're trying to evaluate a limit, not to "solve" a limit. One solves equations; one solves problems; one evaluates expressions.
To evaluate $\lim \limits_{x \to -2} \dfrac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)},$ you don't need to know the zeros of $x^2-3x+5.$ The reason for being concerned with zeros of $x^3 - x^2 - x + 10$ and $x^2 + 3x + 2$ is only that $-2$ is such a zero, so you get $0$ in both the numerator and the denominator. If substitution into polynomials yields a nonzero number in the denominator, then you're done with the "hard" part.
You have
$$
\lim_{x \to -2} \frac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)} = \lim_{x\to-2} \frac{x^2 - 3x+5}{x+1} = \frac{(-2)^2 - 3\cdot(-2) + 5}{-2+1} = \frac {15} {-1} = -15.
$$
A: A very easy way is to use L-Hospital since it's of for $\frac{0}{0}$ therefor your equation will be:
$\frac{3x^2-2x-1}{2x+3}$ now substitue x=-2 and you get answer -15
