Question

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

Answer 1

`lim_(x->2)(sqrt(3x-2)-sqrt(x+2))/(x-2)=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->2)(sqrt(3x-2)-sqrt(x+2))/(x-2)=(sqrt(3(2)-2)-sqrt((2)+2))/((2)-2)=(sqrt(4)-sqrt(4))/0=0/0`

You may be wondering how this helps us. Well, because we have `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 `0/0` or `oo/oo`, 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 `d/dx (sqrt(3x-2)-sqrt(x+2))` here. Using the sum rule, we can simplify this to `d/dxsqrt(3x-2)-d/dxsqrt(x+2)`. Taking the derivative of the first term, we see:
`d/dxsqrt(3x-2)=(3x-2)^(1/2)=3/(2sqrt(3x-2))->`Using power rule and chain rule

For the second term,
`d/dxsqrt(x+2)=(x+2)^(1/2)=1/(2sqrt(x+2))->`Using power rule

Thus, our new numerator is: `3/(2sqrt(3x-2))-1/(2sqrt(x+2))`.

Derivative of Denominator
This one is fairly easy: `d/dx(x-2)=1`. Yep, that's it.

Put it all Together
Combining these two results into one, our new fraction is `(3/(2sqrt(3x-2))-1/(2sqrt(x+2)))/1=3/(2sqrt(3x-2))-1/(2sqrt(x+2))`. We can now evaluate it at `x=2` and see if anything changes:
`lim_(x->2)(3/(2sqrt(3x-2))-1/(2sqrt(x+2)))=3/(2sqrt(3(2)-2))-1/(2sqrt((2)+2))=3/(2sqrt(4))-1/(2sqrt(4))=1/2`

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

Answer 2

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

Explanation:

`(sqrt(3x-2)-sqrt(x+2))/(x-2) = (sqrt(3x-2)-sqrt(x+2))/(x-2) *(sqrt(3x-2)+sqrt(x+2))/(sqrt(3x-2)+sqrt(x+2))`

`= ((3x-2)-(x+2))/((x-2)(sqrt(3x-2)+sqrt(x+2)))`

`= (2x-4)/((x-2)(sqrt(3x-2)+sqrt(x+2)))`

`= (2cancel((x-2)))/(cancel((x-2))(sqrt(3x-2)+sqrt(x+2)))`

So we have

`lim_(xrarr2)(sqrt(3x-2)-sqrt(x+2))/(x-2) = lim_(xrarr2)2/(sqrt(3x-2)+sqrt(x+2))`

`= 2/(sqrt(3(2)-2)+sqrt((2)+2))`

`= 2/(sqrt(4)+sqrt(4)) = 2/(2+2) = 1/2`