# What is the limit of sqrt(x^3-2x^2+1)/[x-1] as x goes to infinity?

Oct 17, 2015

${\lim}_{x \rightarrow \infty} \frac{\sqrt{{x}^{3} - 2 {x}^{2} + 1}}{x - 1} = \infty$

#### Explanation:

$\sqrt{{x}^{3} - 2 {x}^{2} + 1} = \sqrt{{x}^{2} \left(x - 2 + \frac{1}{x} ^ 2\right)}$

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

$= \left\mid x \right\mid \sqrt{x - 2 + \frac{1}{x} ^ 2}$

So

${\lim}_{x \rightarrow \infty} \frac{\sqrt{{x}^{3} - 2 {x}^{2} + 1}}{x - 1} = {\lim}_{x \rightarrow \infty} \frac{\left\mid x \right\mid \sqrt{x - 2 + \frac{1}{x} ^ 2}}{x \left(1 - \frac{1}{x}\right)}$

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

$= \infty$