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}