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