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