In this tutorial we shall discuss an example of limit which involves quadratic functions, and to find the value of this limit we shall factorize the quadratic first and then solve it for the existence of limit.

Let us consider an example which involves a quadratic expression
$\mathop {\lim }\limits_{x \to – 1} \frac{{{x^2} + 6x + 5}}{{{x^2} – 3x – 4}}$

We can see that if we directly apply limit to the denominator of the given function, then the result will be undefined and the limit does not exist. So first we shall factorize the quadratic expression.

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

Now the factors $\left( {x + 1} \right)$ cancel each other out, which causes 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
$\Rightarrow \mathop {\lim }\limits_{x \to – 1} \frac{{{x^2} + 6x + 5}}{{{x^2} – 3x – 4}} = \frac{{\left( { – 1 + 5} \right)}}{{\left( { – 1 – 4} \right)}} = – \frac{4}{5}$