# Example of Limit at Positive Infinity

In this tutorial we shall discuss an example related to the 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 $x > 0$, 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
$\Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{4x + 5}}{{\sqrt {3{x^2} – 7} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{4 + 0}}{{\sqrt {3 – 0} }} = \frac{4}{{\sqrt 3 }}$