Answer to Question #203450 in Calculus for jay

Question

Evaluate lim𝑥→2 𝑥^3−2𝑥^2+4𝑥−8/x^4−2𝑥^3+𝑥−2

Answer 1

$lim_{𝑥→2} \frac{𝑥^3−2𝑥^2+4𝑥−8}{x^4−2𝑥^3+𝑥−2}\hspace{1cm}(\frac{0}{0}form)\\ \text{Apply L'Hospital Rule,}\\ =lim_{𝑥→2} \frac{3𝑥^2−4𝑥+4}{4x^3−6𝑥^2+1}\\ =\frac{8}{9}$