GATE 2019 CS – Question 23
Compute $\displaystyle\lim_{x\to3}\dfrac{x^4-81}{2x^2-5x-3}$
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Correct answer: (C) 108/7
Explanation
Both the numerator and denominator vanish at $x=3$, so apply L'Hôpital's rule: $\frac{4x^3}{4x-5}=\frac{108}{7}$.