Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 61

Determine a number such that $\displaystyle \lim \limits_{x \to -2} \frac{3x^2 + ax + a + 3}{x^2 + x - 2}$ exists. If so, find the value of $a$ and the value of the limit.

The value of the denominator as $x$ approaches -2 will be equal to 0. In order for the limit to exist, we equate the numerator to 0.

$
\begin{equation}
\begin{aligned}
\lim\limits_{x \to -2 } (3x^2 + ax + a + 3) & = 0\\
3 \lim\limits_{x \to -2 } x^2 + a \lim\limits_{x \to -2 } x + \lim\limits_{x \to -2 } a + \lim\limits_{x \to -2 } 3 & = 0\\
3(-2)^2 + a(-2) + a + 3 & = 0\\
12 - 2a + a + 3 & = 0\\
15 - a & = 0\\
a & = 15
\end{aligned}
\end{equation}
$


By substituting the value of $a$, we can now determine the limit..



$
\begin{equation}
\begin{aligned}
\lim \limits_{x \to -2} \frac{3x^2 + ax + a + 3}{x^2 + x - 2} & = \lim \limits_{x \to -2} \frac{3x^2 + 15 x + 15 + 3}{x^2 + x -2}\\
& = \lim \limits_{x \to -2} \frac{(3x + 9)\cancel{(x + 2)}}{\cancel{(x + 2)}(x - 1)}\\
& = \lim \limits_{x \to -2} \frac{3x + 9}{x - 1} \\
& = \frac{3(-2)+9}{-2-1}\\
& = \frac{3}{-3}\\
& = -1

\end{aligned}
\end{equation}
$