24. - / 3.84 points find the number a such that the limit exists. $lim_{x\rightarrow - 2}\frac{3x^{2}+ax + a…

Explanation:

Step1: Factor the denominator

Factor $x^{2}+x - 2=(x + 2)(x - 1)$.

Step2: For the limit to exist

Since the denominator approaches 0 as $x\to - 2$, the numerator must also approach 0 as $x\to - 2$ so that we don't get an infinite - valued limit. Substitute $x=-2$ into the numerator $3x^{2}+ax + a + 3$: $3(-2)^{2}+a(-2)+a + 3=0$. $12-2a+a + 3=0$. $15 - a=0$. $a = 15$.

Step3: Substitute $a = 15$ into the original function

The original function becomes $\lim_{x\to - 2}\frac{3x^{2}+15x+15 + 3}{x^{2}+x - 2}=\lim_{x\to - 2}\frac{3x^{2}+15x+18}{x^{2}+x - 2}$. Factor the numerator: $3x^{2}+15x + 18=3(x^{2}+5x + 6)=3(x + 2)(x+3)$. The function is now $\lim_{x\to - 2}\frac{3(x + 2)(x + 3)}{(x + 2)(x - 1)}$. Cancel out the common factor $(x + 2)$ (since $x\neq - 2$ when taking the limit), we get $\lim_{x\to - 2}\frac{3(x + 3)}{x - 1}$.

Step4: Evaluate the limit

Substitute $x=-2$ into $\frac{3(x + 3)}{x - 1}$: $\frac{3(-2 + 3)}{-2-1}=\frac{3\times1}{-3}=-1$.

Answer:

$a = 15$ $-1$