Question

How do you evaluate the limit of $\lim \left( -{{x}^{3}}+3{{x}^{2}}-4 \right)$ as $x \to -1$?

Answer 1

To solve this question it is important to use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ at first step. After this we will get the respective values of x for equation ${{x}^{2}}-3x$. By this we will have the required factor of it. After this the substitution process will start in which we will substitute x as 3 to get the desired result.

Complete step-by-step answer:
We will start solving this question by first understanding the concept of a limit. By the term limit we mean the closeness of the given function to the given limit. According to this question, we need to find the limit of the function $\lim \left( -{{x}^{3}}+3{{x}^{2}}-4 \right)$ as x come closer to the point – 1. Since, the constant term in this equation is – 4 so, we will check for its factors by substituting points lying between – 4 to 4. So, if we put x = 0 therefore, we get $-{{x}^{3}}+3{{x}^{2}}-4=-{{\left( 0 \right)}^{3}}+3{{\left( 0 \right)}^{2}}-4=-4\ne 0$. But when x = – 1 we get $-{{x}^{3}}+3{{x}^{2}}-4=-{{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}-4=1+3-4=0$. Thus, one of the factors of the given cubic equation is $x+1$. Now, we will divide $-{{x}^{3}}+3{{x}^{2}}-4$ by $x+1$. Therefore, we get

& -{{x}^{3}}+3{{x}^{2}}-4 \\\ & \underline{\mp {{x}^{3}}\mp {{x}^{2}}} \\\ & 4{{x}^{2}}-4 \\\ & \underline{\pm 4{{x}^{2}}\pm 4x} \\\ & -4x-4 \\\ & \underline{\mp 4x\mp 4} \\\ & 0 \\\ \end{aligned}}\right.}}$$ This results into the remaining factors of $-{{x}^{3}}+3{{x}^{2}}-4$ as follows, $\begin{aligned} & \Rightarrow -{{x}^{3}}+3{{x}^{2}}-4=\left( x+1 \right)\left( -{{x}^{2}}+4x-4 \right) \\\ & \Rightarrow -{{x}^{3}}+3{{x}^{2}}-4=\left( x+1 \right)\left( -{{x}^{2}}+2x+2x-4 \right) \\\ & \Rightarrow -{{x}^{3}}+3{{x}^{2}}-4=\left( x+1 \right)\left( x\left( -x+2 \right)-2\left( -x+2 \right) \right) \\\ & \Rightarrow -{{x}^{3}}+3{{x}^{2}}-4=\left( x+1 \right)\left( x-2 \right)\left( -x+2 \right) \\\ \end{aligned}$ Now, we will consider , $\begin{aligned} & \displaystyle \lim_{x \to -1}\left( -{{x}^{3}}+3{{x}^{2}}-4 \right)=\displaystyle \lim_{x \to -1}\left( \left( x+1 \right)\left( x-2 \right)\left( -x+2 \right) \right) \\\ & \Rightarrow \displaystyle \lim_{x \to -1}\left( -{{x}^{3}}+3{{x}^{2}}-4 \right)=\left( -1+1 \right)\left( -1-2 \right)\left( -\left( -1 \right)+2 \right) \\\ & \Rightarrow \displaystyle \lim_{x \to -1}\left( -{{x}^{3}}+3{{x}^{2}}-4 \right)=0\times -3\times 3=0 \\\ \end{aligned}$ Hence, the limit of $\displaystyle \lim_{x \to -1}\left( -{{x}^{3}}+3{{x}^{2}}-4 \right)=0$. **Note:** We could have used the limit directly to the given function such as $\displaystyle \lim_{x \to -1}\left( -{{x}^{3}}+3{{x}^{2}}-4 \right)=-{{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}-4=1+3-4=0$. Also, there is an alternative method for solving $-{{x}^{2}}+4x-4$. That is we can use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and substitute a = - 1, b = 4, c = - 4. This results $-{{x}^{2}}+4x-4$ into, $\begin{aligned} & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( -1 \right)\left( -4 \right)}}{2\left( -1 \right)} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{16-16}}{-2} \\\ & \Rightarrow x=\dfrac{-4\pm 0}{-2} \\\ & \Rightarrow x=2,2 \\\ \end{aligned}$ At last we get the factors as $\left( x-2 \right)\left( x-2 \right)$.