Question

${lim}_{x\to 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}$

Answer 1

${lim}_{x\to 1}\frac{x^4-{3x}^3+2}{x^3-5x^2+3x+1}$

$=\frac{0}{0}$ from

Applying L'hospital Rule,

${lim}_{x\to 1}\frac{\frac{d}{dx}(x^4-{3x}^3+2)}{\frac{d}{dx}(x^3-5x^2+3x+1)}$

$={lim}_{x\to 1}\frac{{4x}^3-9x^2}{3x^2-10x+3}$

$=\frac{4-9}{3-10+3}$

$=\frac{-5}{-}$

$=\frac{5}{4}$

Hence, the answer is $\frac{5}{4}$.