Question Detail

First, check the value of the numerator and denominator as (x \to 3):Numerator: (3^5 - 243 = 243 - 243 = 0).Denominator: (3^3 - 27 = 27 - 27 = 0).Since this is an indeterminate form (0/0), we can apply L'Hôpital's Rule:(\lim_{x\to3} \frac{\frac{d}{dx}(x^5-243)}{\frac{d}{dx}(x^3-27)})(= \lim_{x\to3} \frac{5x^4}{3x^2})(= \lim_{x\to3} \frac{5}{3}x^2)Now, substitute (x = 3):(= \frac{5}{3}(3^2))(= \frac{5}{3}(9))(= 5 \times 3)(= 15).