# What is the limit of (sqrt(x^2+x)-x)  as x approaches infinity?

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Jim H Share
Feb 17, 2016

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + x} - x\right) = \frac{1}{2}$

#### Explanation:

The initial form for the limit is indeterminate $\infty - \infty$

So, use the conjugate.

$\left(\sqrt{{x}^{2} + x} - x\right) = \frac{\sqrt{{x}^{2} + x} - x}{1} \cdot \frac{\sqrt{{x}^{2} + x} + x}{\sqrt{{x}^{2} + x} + x}$

$= \frac{{x}^{2} + x - {x}^{2}}{\sqrt{{x}^{2} + x} + x}$

$= \frac{x}{\sqrt{{x}^{2} + x} + x}$

${\lim}_{x \rightarrow \infty} \frac{x}{\sqrt{{x}^{2} + x} + x}$ has indeterminate form $\frac{\infty}{\infty}$, but we can factor and reduce.

We know that $\sqrt{{x}^{2}} = \left\mid x \right\mid$, so for positive $x$ (which is all we are concerned about for a limit as $x$ increases without bound) we have

$\frac{x}{\sqrt{{x}^{2} + x} + x} = \frac{x}{{\sqrt{x}}^{2} \sqrt{1 + \frac{1}{x}} + x}$ $\text{ }$ (for all $x \ne 0$)

$= \frac{x}{x \sqrt{1 + \frac{1}{x}} + x}$ $\text{ }$ (for $x > 0$)

$= \frac{x}{x \left(\sqrt{1 + \frac{1}{x}} + 1\right)}$

$= \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$

${\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$

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