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..



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