Question

Solve: $\underset{x\to 1}{lim}\frac{x^3-1}{x^4-1}$

Answer 1

$\underset{x\to 1}{lim}\frac{x^3-1}{x^4-1}$ ..... $\frac{0}{0}$ form

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

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

$=\frac{1+1+1}{2\times 2}$

$=\frac{3}{4}$