# What is the limit of ((64x^2 +x)^(1/2) -8x) as x approaches infinity?

May 23, 2016

${\lim}_{x \rightarrow \infty} \left(\sqrt{64 {x}^{2} + x} - 8 x\right) = \frac{1}{16}$

#### Explanation:

${\lim}_{x \rightarrow \infty} \left(\sqrt{64 {x}^{2} + x} - 8 x\right)$ has indeterminate initial form $\infty - \infty$.

We'll write this as a quotient using the conjugate of the expression.

$\left(\sqrt{64 {x}^{2} + x} - 8 x\right) = \frac{\sqrt{64 {x}^{2} + x} - 8 x}{1} \cdot \frac{\left(\sqrt{64 {x}^{2} + x} + 8 x\right)}{\left(\sqrt{64 {x}^{2} + x} + 8 x\right)}$

$= \frac{64 {x}^{2} + x - 64 {x}^{2}}{\sqrt{64 {x}^{2} + x} + 8 x}$

$= \frac{x}{\sqrt{64 {x}^{2} + x} + 8 x}$

For $x \ne 0$,, we can factor ${x}^{2}$ in the radicand.

$= \frac{x}{\sqrt{{x}^{2} \left(64 + \frac{1}{x}\right)} + 8 x}$

Now use $\sqrt{{x}^{2}} = \left\mid x \right\mid$, so for $x > 0$, we get

$= \frac{x}{\sqrt{{x}^{2}} \sqrt{64 + \frac{1}{x}} + 8 x} = \frac{x}{x \sqrt{64 + \frac{1}{x}} + 8 x}$

Simplify:

$= \frac{x}{x \left(\sqrt{64 + \frac{1}{x}} + 8\right)} = \frac{1}{\sqrt{64 + \frac{1}{x}} + 8}$ (for $x > 0$)

Finally, evaluate the limit.

${\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{64 + \frac{1}{x}} + 8} = \frac{1}{\sqrt{64 + 0} + 8} = \frac{1}{16}$