Question

Find the following limit.
$\underset{x\to 1}{lim}\frac{x^4-2x^2+1}{x^3-1}.$

Answer 1

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

By using factorization method

$\underset{x\to 1}{lim}\frac{(x^2-1{)}^2}{(x-1)(x^2+x+1)}$ [here $(x^2-1{)}^2=x^4-2x^2-1x^3-1^3=(x-1\left)\right(x^2+x+1\left)\right]$

$\underset{x\to 1}{lim}\frac{\left[\right(x+1\left)\right(x-1){]}^2}{(x-1)(x^2+x+1)}$

$\underset{x\to 1}{lim}\frac{(x+1{)}^2(x-1{)}^2}{(x-1)(x^2+x+1)}$

$\underset{x\to 1}{lim}\frac{(x+1{)}^2(x-1)}{x^2+x+1}$

$=\frac{(1+1{)}^2(0)}{1+1+1}=0$.