# How do you evaluate the limit of lim (-x^3+3x^2-4) as x->-1?

Jun 13, 2017

$\therefore x \rightarrow \left(- 1\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = 0$

#### Explanation:

$: x \rightarrow \left(- 0.9999\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = - 0.0009$

$x \rightarrow \left(- {1}^{+}\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = \lim \left(- {\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} - 4\right) = 0$

$: x \rightarrow \left(- 1.0001\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = 0.0009$

$x \rightarrow \left(- {1}^{-}\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = \lim \left(- {\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} - 4\right) = 0$

$\therefore x \rightarrow \left(- 1\right) \lim \left(- {x}^{3} + 3 {x}^{2} - 4\right) = 0$ [Ans]

Jun 13, 2017

$0$

#### Explanation:

$\text{the limit of a polynomial can be evaluated by substitution}$

$\Rightarrow {\lim}_{x \to - 1} \left(- {x}^{3} + 3 {x}^{2} - 4\right)$

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

$= 0$