Question

(a) find a number n such that lim(x→ - 2) (3x² + nx + n + 3)/(x² + x - 2) exists. (b) after finding n, what is the value of the limit?

Explanation:

Step1: Factor the denominator

Factor $x^{2}+x - 2=(x + 2)(x-1)$. For the limit $\lim_{x
ightarrow - 2}\frac{3x^{2}+nx + n + 3}{x^{2}+x - 2}$ to exist, $x + 2$ must be a factor of the numerator $3x^{2}+nx + n + 3$.

Step2: Use the factor - theorem

If $x+2$ is a factor of $3x^{2}+nx + n + 3$, then when $x=-2$, $3x^{2}+nx + n + 3 = 0$. Substitute $x=-2$ into $3x^{2}+nx + n + 3$:
\[

$$\begin{align*} 3\times(-2)^{2}+n\times(-2)+n + 3&=0\\ 3\times4-2n + n+3&=0\\ 12 - 2n + n+3&=0\\ 15 - n&=0\\ n&=15 \end{align*}$$

\]

Step3: Rewrite the rational function

Substitute $n = 15$ into the original function, we get $\frac{3x^{2}+15x+15 + 3}{x^{2}+x - 2}=\frac{3x^{2}+15x+18}{x^{2}+x - 2}=\frac{3(x^{2}+5x + 6)}{(x + 2)(x - 1)}=\frac{3(x + 2)(x+3)}{(x + 2)(x - 1)}$ for $x
eq - 2$.

Step4: Simplify the rational function and find the limit

Cancel out the common factor $(x + 2)$ (since $x
ightarrow - 2$ but $x
eq - 2$), we have $\lim_{x
ightarrow - 2}\frac{3(x + 2)(x + 3)}{(x + 2)(x - 1)}=\lim_{x
ightarrow - 2}\frac{3(x + 3)}{x - 1}$. Then substitute $x=-2$ into $\frac{3(x + 3)}{x - 1}$:
\[
\frac{3(-2 + 3)}{-2-1}=\frac{3\times1}{-3}=-1
\]

Answer:

(a) $15$
(b) $-1$