# How do you find the limit of  (sqrt(3x - 2) - sqrt(x + 2))/(x-2) as x approaches 2?

Mar 11, 2016

${\lim}_{x \to 2} \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = \frac{1}{2}$

#### Explanation:

The first thing we do in limit problems is substitute the $x$ value in question and see what happens. Using $x = 2$, we find:
${\lim}_{x \to 2} \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2} = \frac{\sqrt{3 \left(2\right) - 2} - \sqrt{\left(2\right) + 2}}{\left(2\right) - 2} = \frac{\sqrt{4} - \sqrt{4}}{0} = \frac{0}{0}$

You may be wondering how this helps us. Well, because we have $\frac{0}{0}$, this problem becomes fair game for an application of L'Hopital's Rule. This rule says that if we evaluate a limit and get $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can find the derivative of the numerator and denominator and try evaluating it then.

So, without further ado, let's get to it.

Derivative of Numerator
We're trying to find $\frac{d}{\mathrm{dx}} \left(\sqrt{3 x - 2} - \sqrt{x + 2}\right)$ here. Using the sum rule, we can simplify this to $\frac{d}{\mathrm{dx}} \sqrt{3 x - 2} - \frac{d}{\mathrm{dx}} \sqrt{x + 2}$. Taking the derivative of the first term, we see:
$\frac{d}{\mathrm{dx}} \sqrt{3 x - 2} = {\left(3 x - 2\right)}^{\frac{1}{2}} = \frac{3}{2 \sqrt{3 x - 2}} \to$Using power rule and chain rule

For the second term,
$\frac{d}{\mathrm{dx}} \sqrt{x + 2} = {\left(x + 2\right)}^{\frac{1}{2}} = \frac{1}{2 \sqrt{x + 2}} \to$Using power rule

Thus, our new numerator is: $\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}$.

Derivative of Denominator
This one is fairly easy: $\frac{d}{\mathrm{dx}} \left(x - 2\right) = 1$. Yep, that's it.

Put it all Together
Combining these two results into one, our new fraction is $\frac{\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}}{1} = \frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}$. We can now evaluate it at $x = 2$ and see if anything changes:
${\lim}_{x \to 2} \left(\frac{3}{2 \sqrt{3 x - 2}} - \frac{1}{2 \sqrt{x + 2}}\right) = \frac{3}{2 \sqrt{3 \left(2\right) - 2}} - \frac{1}{2 \sqrt{\left(2\right) + 2}} = \frac{3}{2 \sqrt{4}} - \frac{1}{2 \sqrt{4}} = \frac{1}{2}$

And there you have it! We can confirm this result by looking at the graph of $\frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{x - 2}$:
graph{(sqrt(3x-2)-sqrt(x+2))/(x-2) [-0.034, 3.385, -0.205, 1.504]}

Mar 14, 2016

If your calculus course has not yet covered derivatives and l'Hospital's rule, use algebra to rationalize the numerator.

#### Explanation:

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

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

$= \frac{2 x - 4}{\left(x - 2\right) \left(\sqrt{3 x - 2} + \sqrt{x + 2}\right)}$

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

So we have

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

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

$= \frac{2}{\sqrt{4} + \sqrt{4}} = \frac{2}{2 + 2} = \frac{1}{2}$