Limits at Positive Infinity with Radicals

In this tutorial we shall discuss an example related to the limit at positive 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} + 3x} - x} \right)

By rationalizing, we have

\begin{gathered} \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) \times \left( {\frac{{\sqrt {{x^2} + 3x} + x}}{{\sqrt {{x^2} + 3x} + x}}} \right) \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\left( {\sqrt {{x^2} + 3x} } \right)}^2} - {{\left( x \right)}^2}}}{{\sqrt {{x^2} + 3x} + x}} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + 3x - {x^2}}}{{\sqrt {{x^2} + 3x} + x}} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{3x}}{{\sqrt {{x^2} + 3x} + 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} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{3x}}{{\left| x \right|}}}}{{\frac{{\sqrt {{x^2} + 3x} + 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} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{3x}}{x}}}{{\frac{{\sqrt {{x^2} + 3x} }}{{\sqrt {{x^2}} }} + \frac{x}{x}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{\sqrt {\frac{{{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}}} + 1}} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{\sqrt {1 + \frac{3}{x}} + 1}} \\ \end{gathered}

By applying limits, we have

 \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{\sqrt {1 + 0} + 1}} = \frac{3}{2}