If $L = lim x \to - 2 \frac{3 x^{2} + k x + k - 7}{2 x^{2} + x - 6}$ exists and is finite, where $k$ is a real number, then the values of $k$ and $L$ are

k = 5, L = - 1

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k = 5, L = 1

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k \in R, L = \frac{3}{2}

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k = - 5, L = 1

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Solution

The correct option is B $k = 5, L = 1$
$L = lim x \to - 2 \frac{3 x^{2} + k x + k - 7}{2 x^{2} + x - 6}$
For $\frac{0}{0}$ form and limit to exist, numerator should tend to $0$ as $x \to - 2$
$3 (- 2)^{2} + k (- 2) + k - 7 = 0 \Rightarrow k = 5 L = lim x \to - 2 \frac{3 x^{2} + 5 x - 2}{2 x^{2} + x - 6} = lim x \to - 2 \frac{(3 x - 1) (x + 2)}{(2 x - 3) (x + 2)} = \frac{- 6 - 1}{- 4 - 3} = 1$

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