Question

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

Answer 1

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

Putting x =1 in the numerator and denominator the expression $\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}$ takes $\left(\frac{0}{0}\right)$ form, which means (x-1)in a factor of both numerator and denominator. So, for factorising divide both numerator and denominator by x -1 as follows

$\therefore x^4-3x^3+2=(x-1)(x^3-2x^2-2x-2)$ by division algorithm

Again

$\therefore x^3-5x^2+3x+1=(x-1)(x^2-4x-1)$ by division algorithm

$\therefore {lim}_{x\to 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}$ [It is of $\left(\frac{0}{0}\right)$ form]

$={lim}_{x\to 1}\frac{(x-1)(x^3-2x^2-2x-2)}{(x-1)(x^2-4x-1)}$

$={lim}_{x\to 1}\frac{(x^3-2x^2-2x-2)}{(x^2-4x-1)}$

$=\frac{(1{)}^3-2(1{)}^2-2\left(1\right)-2}{(1{)}^2-4(1)-1}$

$=\frac{1-2-2-2}{1-4-1}=\frac{1-6}{1-5}=\frac{-5}{-4}=\frac{5}{4}$