Limit by Factoring Cubic Expression

In this tutorial we shall discuss an example of evaluating limits involving cubic expression. In most cases, if a limit involves cubic expressions we can factorize by using the method of synthetic division.

Let us consider an example which involve cubic expression:

\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - 5x + 3}}{{{x^3} - 3x + 2}}

We observe that if we directly apply limit to the denominator of the given function, then the result will be undefined and limit does not exist. So first we shall factorize by using synthetic division.

We have

\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - 5x + 3}}{{{x^3} - 3x + 2}}


Consider the expression {x^3} + {x^2} - 5x + 3, using synthetic division, we have

\begin{gathered} 1\left| \!{ {\, {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1&1&{ - 5}&3 \end{array}} \\ {\begin{array}{*{20}{c}} {}&1&2&{ - 3} \end{array}} \end{array}} \,}} \right. \\ \,\,\,\,\begin{array}{*{20}{c}} 1&2&{ - 3}&0 \end{array} \\ \end{gathered}


\therefore {x^3} + {x^2} - 5x + 3 = \left( {x - 1} \right)\left( {{x^2} + 2x - 3} \right) = {\left( {x - 1} \right)^2}\left( {x + 3} \right)

Now consider the expression {x^3} - 3x + 2. Using synthetic division, we have

\begin{gathered} 1\left| \!{ {\, {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1&0&{ - 3}&2 \end{array}} \\ {\begin{array}{*{20}{c}} {}&1&1&{ - 2} \end{array}} \end{array}} \,}} \right. \\ \,\,\,\,\begin{array}{*{20}{c}} 1&1&{- 2}&0 \end{array} \\ \end{gathered}


\therefore {x^3} - 3x + 2 = \left( {x - 1} \right)\left( {{x^2} + x - 2} \right) = {\left( {x - 1} \right)^2}\left( {x + 2} \right)

Now we have the limit

\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - 5x + 3}}{{{x^3} - 3x + 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {x - 1} \right)}^2}\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}

Now the terms {\left( {x - 1} \right)^2} cancel each other out, which causes the limit to be undefined

\Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - 5x + 3}}{{{x^3} - 3x + 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 3} \right)}}{{\left( {x + 2} \right)}}

Applying the limits, we have

\Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - 5x + 3}}{{{x^3} - 3x + 2}} = \frac{{\left( {1 + 3} \right)}}{{\left( {1 + 2} \right)}} = \frac{4}{3}