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

Apr 22, 2016

Don't think too much about what you should do to solve the problem. Think about what you could do and ask if that helps.

#### Explanation:

Because the expression is not real-valued left of $2$, a beginning calculus course will say there is no (two-sided) limit as $x$ approaches $2$. We can find the limit as $x$ approaches $2$ from the right. (Thanks George C., for pointing this out.)

$\frac{\sqrt{x} - \sqrt{2}}{\sqrt{{x}^{2} - 2 x}}$ as $x \rightarrow {2}^{+}$, this expression has indeterminate form $\frac{0}{0}$.

${x}^{2} - 2 x$ is not a perfect square. I could complete the square to get ${x}^{2} - 2 x + 1 - 1 = {\left(x - 1\right)}^{2} - 1$. That's not a perfect square. It is a difference of squares, which might be helpful for some problems, but I don't see a way to use it here.

We could write $\sqrt{{x}^{2} - 2 x} = \sqrt{{x}^{2}} \sqrt{1 - \frac{2}{x}}$ $\text{ }$ (for $x \ne 0$). That doesn't look promising.

We can rationalize the numerator by multiplying by $\frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}$.
We get

$\frac{x - 2}{\sqrt{{x}^{2} - 2 x} \left(\sqrt{x} + \sqrt{2}\right)}$ The form of the limit as $x \rightarrow {2}^{+}$ is still indeterminate, but wait there's more I could do now. I have something involving $x - 2$ in the denominator.

Factor $\sqrt{x}$ out of $\sqrt{{x}^{2} - 2 x}$

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

Now reduce the fraction using $\frac{a}{\sqrt{a}} = {\left(\sqrt{a}\right)}^{2} / \sqrt{a} = \frac{\cancel{\sqrt{a}} \sqrt{a}}{\cancel{\sqrt{a}}} = \sqrt{a}$

$\frac{x - 2}{\sqrt{x} \sqrt{x - 2} \left(\sqrt{x} + \sqrt{2}\right)} = \frac{\sqrt{x - 2}}{\sqrt{x} \left(\sqrt{x} + \sqrt{2}\right)}$.

The limit as $x \rightarrow {2}^{+}$ is $\frac{0}{\sqrt{2} \left(\sqrt{2} + \sqrt{2}\right)} = 0$.

Presenting it without the false starts and explanation looks like:

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

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

$= \frac{0}{\sqrt{2} \left(\sqrt{2} + \sqrt{2}\right)} = 0$