Limit by Factoring Cubic Expression

In this tutorial we shall discuss an example to evaluating limits involving cubic expression, in most of the cases if limit involves cubic expression and we can factorize by using 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 directly limit in the denominator of the given function, then the result will be undefined and limit will does not exist, so first we shall factorize by using synthetic division, we have
Now

\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 which cause 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}

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