Question

If $li{m}_{x\to -3}\frac{3x^2+ax+a-7}{x^2+2x-3}$, exists, then $a=?$

• A. $10$
• B. $15$
• C. $-15$
• D. $-10$

Answer 1

The correct option is A

$10$

Explanation for the correct option.

$\begin{array}{rcl}{x}^2+2x-3& =& x^2-x+3x-3\\ & =& \left(x-1\right)\left(x+3\right)\end{array}$

So, the denominator tends to $0$, for $x=-3$, thus the numerator should also be equal to $0$ as $li{m}_{x\to -3}\frac{3x^2+ax+a-7}{x^2+2x-3}$ exists.

$\begin{array}{rcl}& \Rightarrow & \underset{x\to -3}{lim}3x^2+ax+a-7=0\\ & \Rightarrow & 3{\left(-3\right)}^2+a\left(-3\right)+a-7=0\\ \Rightarrow 27-3a+a-7& =& 0\\ \Rightarrow 2a& =& 20\\ \Rightarrow a& =& 10\end{array}$

Hence, option A is correct.