Question

como calcular o limite 2x^3- 250 sobre x^2-6x +5 com x tendendo a 5

Answer 1

$\lim_{x \to 5} \dfrac{2x^3-250}{x^2-6x+5} = \lim_{x \to 5} \dfrac{2 (x^3-125)}{(x-5)(x -1)} \Rightarrow \\\\ \lim_{x \to 5} \dfrac{2(x^3 - 5^3)}{(x-5)(x-1)} = \lim_{x \to 5} \dfrac{2(x-5)(x^2+5x+25)}{(x-5)(x-1)} \Rightarrow \\\\ \lim_{x \to 5} \dfrac{2 (x^2+5x+25)}{x-1} =\dfrac{2 (5^2+5.5+25)}{5-1}= \dfrac{150}{4} = \dfrac{75}{2}$