Limits at Positive Infinity with Radicals

In this tutorial we shall discuss an example relating with limit at positive infinity with radial form of 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, so 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}

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