Limit with a twist

Calculus Level 2

lim ⁡ x → 5 x 2 + k x − 27 k + 19 x 2 − 3 x − 10 \lim_{x\to 5}\frac{x^2+kx-27k+19}{x^2-3x-10}

Find k k so that the limit exists, and then evaluate the limit. Submit the limit as answer.

12 7 \frac {12}7 4 4 13 6 \frac {13}6 9 5 \frac 95

S. P. by S. P. , 16, USA

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1 solution

For the limit to exist, the numerator must be zero (since the denominator is zero in the given limit). So 25 + 5 k − 27 k + 19 = 0 25+5k-27k+19=0 or k = 2 k=\boxed 2 . Then the limiting value of the given expression is 2 × 5 + 2 2 × 5 − 3 = 12 7 \dfrac{2\times 5+2}{2\times 5-3}=\boxed {\dfrac{12}{7}} (applying L'Hospital's rule).

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