Question

If $f\left(x\right)=\frac{3x^2+ax+a+1}{x^2+x-2}$, then which of the following can be correct?

• A. $\underset{x\to 1}{lim}f\left(x\right)$ exists $\Rightarrow a=-2$
• B. $\underset{x\to -2}{lim}f\left(x\right)$ exists $\Rightarrow a=13$
• C. $\underset{x\to 1}{lim}f\left(x\right)=\frac{4}{3}$
• D. $\underset{x\to -2}{lim}f\left(x\right)=-\frac{1}{3}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\underset{x\to 1}{lim}f\left(x\right)$ exists $\Rightarrow a=-2$
B $\underset{x\to -2}{lim}f\left(x\right)$ exists $\Rightarrow a=13$
C $\underset{x\to 1}{lim}f\left(x\right)=\frac{4}{3}$
D $\underset{x\to -2}{lim}f\left(x\right)=-\frac{1}{3}$

$f\left(x\right)=\frac{3x^2+ax+a+1}{(x+2)(x-1)}$
As $x\to 1,$ Denominator $\to 0$. Hence as $x\to 1,$ Numerator $\to 0$.
Therefore, $3+2a+1=0$ or $a=-2$
As $x\to -2,$Denominator $\to 0$.Hence as $x\to -2,$ Numerator $\to 0$.
Therefore, $12-2a+a+1=0$ or $a=13$
Now, ${lim}_{x\to 1}f\left(x\right)=\underset{x\to 1}{lim}\frac{3x^2-2x-1}{(x+2)(x-1)}$
$=\underset{x\to 1}{lim}\frac{(3x+1)(x-1)}{(x+2)(x-1)}=\frac{4}{3}$
Now,$\underset{x\to -2}{lim}\frac{3x^2+13x+14}{(x+2)(x-1)}$
$=\underset{x\to -2}{lim}\frac{(3x+7)(x+2)}{(x+2)(x-1)}=-\frac{1}{3}$