Question

Evaluate the limit : $\underset{x\to -1}{lim}\frac{x^3-3x+1}{x-1}$

Answer 1

Using Direct substitution method,

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

$=\frac{-1+3+1}{-2}=\frac{-3}{2}$

Therefore,

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

Hence, the value of limit

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