# Example of Limit at Negative Infinity

In this tutorial we shall discuss an example related to the limit of a function at negative infinity, i.e. $x \to – \infty$.

Let us consider an example:
$\mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }}$

We divide the numerator and denominator of the fraction by $\left| x \right|$. Since we are considering only negative values of $x$ and $\sqrt {{x^2}} = \left| x \right| = – x$ for $x < 0$, using these values we have
$\mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{5x + 6}}{{\left| x \right|}}}}{{\frac{{\sqrt {4{x^2} – 8} }}{{\left| x \right|}}}}$

Using the relation $\sqrt {{x^2}} = \left| x \right| = – x$, we have
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{5x + 6}}{{ – x}}}}{{\frac{{\sqrt {4{x^2} – 8} }}{{\sqrt {{x^2}} }}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{5x}}{{ – x}} + \frac{6}{{ – x}}}}{{\sqrt {\frac{{4{x^2} – 8}}{{{x^2}}}} }} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 5 – \frac{6}{x}}}{{\sqrt {4 – \frac{8}{{{x^2}}}} }} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \frac{{ – 5 – \mathop {\lim }\limits_{x \to – \infty } \frac{6}{x}}}{{\sqrt {4 – \mathop {\lim }\limits_{x \to – \infty } \frac{8}{{{x^2}}}} }} \\ \end{gathered}$

By applying limits, we have
$\Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \frac{{ – 5 – 0}}{{\sqrt {4 – 0} }} = – \frac{5}{2}$