Example of Limit at Positive Infinity

In this tutorial we shall discuss an example relating with limit of a function at positive infinity, i.e. x \to + \infty .

Let us consider an example

\mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} - 7} }}

We divide the numerator and denominator of the fraction by \left| x \right|. Since we are considering only positive 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{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{4x + 5}}{{\left| x \right|}}}}{{\frac{{\sqrt {3{x^2} - 7} }}{{\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{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{4x + 5}}{x}}}{{\frac{{\sqrt {3{x^2} - 7} }}{{\sqrt {{x^2}} }}}} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{4x}}{x} + \frac{5}{x}}}{{\sqrt {\frac{{3{x^2} - 7}}{{{x^2}}}} }} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{4 + \frac{5}{x}}}{{\sqrt {3 - \frac{7}{{{x^2}}}} }} \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \frac{{4 + \mathop {\lim }\limits_{x \to + \infty } \frac{5}{x}}}{{\sqrt {3 - \mathop {\lim }\limits_{x \to + \infty } \frac{7}{{{x^2}}}} }} \\ \end{gathered}

By applying limits, we have

\Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} - 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{4 + 0}}{{\sqrt {3 - 0} }} = \frac{4}{{\sqrt 3 }}