Example of Limit at Negative Infinity

In this tutorial we shall discuss an example relating with 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, so 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}

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