Question

Let $L=\underset{x\to -2}{lim}\frac{3x^2+ax+a+1}{x^2+x-2}.$ If $L$ is finite, then the value of $a+18L$ is

• A. $\frac{1}{3}$
• B. $13$
• C. $7$
• D. $-\frac{1}{3}$

Answer 1

Answer: (C)

$7$

$L=\underset{x\to -2}{lim}\frac{3x^2+ax+a+1}{x^2+x-2}.$
As $x\to -2$, the denominator tends to $0$, so for existence of limit, the numerator should also tend to $0$.
$\Rightarrow \underset{x\to -2}{lim}3x^2+ax+a+1=0$
$\Rightarrow 13-a=0$
$\Rightarrow a=13$

$L=\underset{x\to -2}{lim}\frac{3x^2+13x+14}{x^2+x-2}$
It is of the form $\frac{0}{0}$, so applying L-hospital's rule, we get

$L=\underset{x\to -2}{lim}\frac{6x+13}{2x+1}$
$L=-\frac{1}{3}$
So, $a+18L=7$