Question

Show that
$\lim \limits_{x \to \infty}( \sqrt{ {x}^{2} + x } -x) = \dfrac{1}{2}$
​

Answer 1

$\Large\text{\underline{\underline{Important!}}}$

The rational function $f(x)=\dfrac{1}{x}$ approaches one of the asymptotes, the x-axis.

$\text{ \cdots\longrightarrow\underline{\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0.} }$

Here, this limit finds the limiting value.

$\Large\text{\underline{\underline{Explanation}}}$

As, -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\sqrt{x^{2}+x}=\lim_{x\to\infty}x\sqrt{1+\dfrac{1}{x}}=\boxed{\infty.} }$

$\cdots\longrightarrow\displaystyle\lim_{x\to\infty}-x=\boxed{-\infty.}$

It is in the form of, -

$\text{ \cdots\longrightarrow\infty-\infty , which is indeterminate.}$

But we can not say the limit does not exist for indeterminate form. Let us recall the polynomial identity -

$\text{ \cdots\longrightarrow\underline{a^{2}-b^{2}=(a+b)(a-b).} }$

We are going to find the expression before rationalization. Then, -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&(\sqrt{x^{2}+x}-x)\times\dfrac{\sqrt{x^{2}+x}+x}{\sqrt{x^{2}+x}+x}\\\\&=\dfrac{(\sqrt{x^{2}+x})^{2}-x^{2}}{\sqrt{x^{2}+x}+x}\\\\&=\dfrac{(x^{2}+x)-x^{2}}{\sqrt{x^{2}+x}+x}\\\\&=\dfrac{x}{\sqrt{x^{2}+x}+x}.\end{aligned}} }$

As a solution to find the limiting value, we divide the numerator and denominator. Then, -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&\dfrac{x}{\sqrt{x^{2}+x}+x}\\\\&=\dfrac{\frac{1}{x}(x)}{\frac{1}{x}(\sqrt{x^{2}+x}+x)}\\\\&=\dfrac{1}{\sqrt{\frac{x^{2}+x}{x^{2}}}+1}\\\\&=\dfrac{1}{\sqrt{1+\frac{1}{x}}+1}.\end{aligned}} }$

As, -

$\text{ \cdots\longrightarrow\boxed{\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0.} }$

Finally, we can apply the limit which is, -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\frac{1}{x}}+1}=\dfrac{1}{2}. }$

So, the answer is -

$\text{ \cdots\longrightarrow\boxed{\displaystyle\lim_{x\to\infty}(\sqrt{x^{2}+x}-x)=\dfrac{1}{2}.} }$