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

Oct 17, 2015

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

#### Explanation:

$\sqrt{{x}^{2} - 2 x + 3 - \sqrt{x}} = \sqrt{{x}^{2}} \sqrt{1 - \frac{2}{x} + \frac{3}{x} ^ 2 = \frac{\sqrt{x}}{x} ^ 2}$

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

As $x \rightarrow \infty$, the radicand goes to $1$, while $\left\mid x \right\mid = x \rightarrow \infty$

Indeed, $y = x$ is an oblique asymptote on the right for the graph of $f \left(x\right) = \sqrt{{x}^{2} - 2 x + 3 - \sqrt{x}}$.