# What is the limit of sqrt(10x^2 -7x +6)/(5x + 8) as x approaches infinity?

Feb 17, 2016

$\sqrt{\frac{2}{5}}$

#### Explanation:

In the bottom fraction $5 x + 8$ will start to look more like $5 x$ as $x$ gets very big.

At the same time $\sqrt{10 {x}^{2} - 7 x + 6}$ will start to look like $\sqrt{10 {x}^{2}} = \sqrt{10} x$ as the ${x}^{2}$ term dominates the others.

Thus $\frac{\sqrt{10 {x}^{2} - 7 x + 6}}{5 x + 8} \to \frac{\sqrt{10} x}{5 x}$ as $x \to \infty$

$\frac{\sqrt{10} x}{5 x} = \frac{\sqrt{10}}{5} = \frac{\sqrt{2} \sqrt{5}}{5} = \sqrt{\frac{2}{5}}$

Hence our final answer: Here is a graph of the function. As you can see it asymptotically approaches $\sqrt{\frac{2}{5}}$

We also see that as $x$ goes to negative infinity the function tends towards $- \sqrt{\frac{2}{5}}$.

Feb 17, 2016

${\lim}_{x \rightarrow \infty} \frac{\sqrt{10 {x}^{2} - 7 x + 6}}{5 x + 8} = \frac{\sqrt{10}}{5}$

#### Explanation:

For all $x \ne 0$, we have

$\sqrt{10 {x}^{2} - 7 x + 6} = \sqrt{{x}^{2}} \sqrt{10 - \frac{7}{x} + \frac{6}{x} ^ 2}$

Furthermore, $\sqrt{{x}^{2}} = \left\mid x \right\mid$.

When looking for a limit as $x$ increses without bound, we are concerned only with positive values of $x$, so

${\lim}_{x \rightarrow \infty} \frac{\sqrt{10 {x}^{2} - 7 x + 6}}{5 x + 8} = {\lim}_{x \rightarrow \infty} \frac{\sqrt{{x}^{2}} \sqrt{10 - \frac{7}{x} + \frac{6}{x} ^ 2}}{x \left(5 + \frac{8}{x}\right)}$

$= {\lim}_{x \rightarrow \infty} \frac{\cancel{x} \sqrt{10 - \frac{7}{x} + \frac{6}{x} ^ 2}}{\cancel{x} \left(5 + \frac{8}{x}\right)}$

$= \frac{\sqrt{10}}{5}$

Bonus

If we look for the limit as $x$ decreases without bound, then we are concerned with negative values of $x$. For those values we have

For $x < 0$, $\sqrt{{x}^{2}} = \left\mid x \right\mid = - x$.

Therefore,

${\lim}_{x \rightarrow - \infty} \frac{\sqrt{10 {x}^{2} - 7 x + 6}}{5 x + 8} = {\lim}_{x \rightarrow - \infty} \frac{\sqrt{{x}^{2}} \sqrt{10 - \frac{7}{x} + \frac{6}{x} ^ 2}}{x \left(5 + \frac{8}{x}\right)}$

$= {\lim}_{x \rightarrow - \infty} \frac{- \cancel{x} \sqrt{10 - \frac{7}{x} + \frac{6}{x} ^ 2}}{\cancel{x} \left(5 + \frac{8}{x}\right)}$

$= - \frac{\sqrt{10}}{5}$