Question

Evaluate the limit:

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

Answer 1

Given: $\underset{x\to 1}{lim}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}$ becomes $\frac{0}{0}$ form

Now numerator,

$\Rightarrow x^4-3x^3+2=(x-1)(x^3-2x^2-2x-2)$

Also, denominator

$\Rightarrow x^3-5x^2+3x+1=(x-1)(x^2-4x-1)$

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

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

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

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

$=\frac{(1-2-2-2)}{(1-4-1)}=\frac{5}{4}$

$Therefore,$

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