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
\[ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{5x + 6}}{{\sqrt {4{x^2} – 8} }} = \frac{{ – 5 – 0}}{{\sqrt {4 – 0} }} = – \frac{5}{2}\]