Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-3\text{x}+1}{\text{x}-1}$

Answer

$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-3\text{x}+1}{\text{x}-1}=\frac{(-1)^3-3(-1)+1}{(-1-1)}$ $=\frac{-1+3+1}{-2}=\frac{3}{-2}=\frac{-3}{2}$

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