# Limits at Negative Infinity with Radicals

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

Let us consider an example:
$\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right)$

By rationalizing, we have
$\begin{gathered} \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) \times \left( {\frac{{\sqrt {{x^2} + 6x} – x}}{{\sqrt {{x^2} + 6x} – x}}} \right) \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{{{\left( {\sqrt {{x^2} + 6x} } \right)}^2} – {{\left( x \right)}^2}}}{{\sqrt {{x^2} + 6x} – x}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{{x^2} + 6x – {x^2}}}{{\sqrt {{x^2} + 6x} – x}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{6x}}{{\sqrt {{x^2} + 6x} – x}} \\ \end{gathered}$

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
$\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{6x}}{{\left| x \right|}}}}{{\frac{{\sqrt {{x^2} + 6x} – x}}{{\left| x \right|}}}}$

Using the relation $\sqrt {{x^2}} = \left| x \right| = – x$, we have
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{6x}}{{ – x}}}}{{\frac{{\sqrt {{x^2} + 6x} }}{{\sqrt {{x^2}} }} – \frac{x}{{ – x}}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 6}}{{\frac{{\sqrt {{x^2} + 6x} }}{{\sqrt {{x^2}} }} + 1}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 6}}{{\sqrt {\frac{{{x^2}}}{{{x^2}}} + \frac{{6x}}{{{x^2}}}} + 1}} \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 6}}{{\sqrt {1 + \frac{6}{x}} + 1}} \\ \end{gathered}$

By applying limits, we have
$\Rightarrow \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} + 6x} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 6}}{{\sqrt {1 + 0} + 1}} = – \frac{6}{2} = – 3$