Limit Involving Quadratic Functions

In this tutorial we shall discuss an example of limit which involves quadratic function, and to find the value of limit we shall factorize the quadratic first and then solve it for the existence of limit.
Let us consider an example which involve quadratic expression

\mathop  {\lim }\limits_{x \to - 1} \frac{{{x^2}  + 6x + 5}}{{{x^2} - 3x - 4}}


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 the quadratic expression, we have
Now

\begin{gathered} \mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 6x + 5}}{{{x^2} - 3x -  4}} \\ = \mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 5x + x + 5}}{{{x^2} - 4x  + x - 4}} \\ = \mathop {\lim }\limits_{x \to - 1} \frac{{x\left( {x + 5} \right) + 1\left(  {x + 5} \right)}}{{x\left( {x - 4} \right) + 1\left( {x - 4} \right)}} \\ = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 5} \right)\left( {x +  1} \right)}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} \\ \end{gathered}


Now the factor \left( {x + 1} \right) cancel each other which cause the limit to be undefined

  \Rightarrow \mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + 6x + 5}}{{{x^2} - 3x - 4}} = \mathop {\lim  }\limits_{x \to - 1} \frac{{\left( {x +  5} \right)}}{{\left( {x - 4} \right)}}


Applying the limits, we have

\Rightarrow  \mathop {\lim }\limits_{x \to - 1}  \frac{{{x^2} + 6x + 5}}{{{x^2} - 3x - 4}} = \frac{{\left( { - 1 + 5}  \right)}}{{\left( { - 1 - 4} \right)}} =  - \frac{4}{5}

Comments

comments