Limits at Negative Infinity with Radicals

In this tutorial we shall discuss an example relating with limit at negative infinity with radial form of 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, so 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

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