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 }}

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