# How do you find the limit of sqrt(x-2) /sqrt( x^2 - 4x -4) as x approaches 2?

Sep 8, 2016

The limit does not exist.

#### Explanation:

We have two square root functions. The restriction that exists for the typical square root function $f \left(x\right) = \sqrt{x}$ is that $x \ge 0$.

So, we know that both $x - 2 \ge 0$ and ${x}^{2} - 4 x - 4 \ge 0$ for this function to be defined.

The first can be easily manipulated to show that $x \ge 2$. Since we want the limit as $x$ approaches $x = 2$, this isn't a problem yet.

To solve the quadratic inequality, first consider ${x}^{2} - 4 x - 4 = 0$. This is solvable through the quadratic formula or completing the square to show that this quadratic has no real solutions, thus it never crosses the $x$-axis. Testing any $x$-value shows that this function will always lie under the $x$-axis and never is greater than or equal to $0$.

Thus, due to the function $\frac{\sqrt{x - 2}}{\sqrt{{x}^{2} - 4 x - 4}}$ not existing at $x = 2$, or any other point, the limit does not exist.

Please comment if you meant to say $\frac{\sqrt{x - 2}}{\sqrt{{x}^{2} - 4 x + 4}}$ instead.