I'm so lost. We're working on limits for this assignment and I get that, but my difficulty lies in breaking down the original problem.

The problem is:

Find the limit:

lim (x->5)

The numerator is (the square root of 9-x) -2 and the denominator is x-(the square root of x+20).

I know that I can't substitute 5 for x because it'll create 0/0, but other than making a table of values close to 5, I don't know what to do. My teacher says to rationalize the numerator and the denominator at the same time, but I have no clue as to what he's talking about.

Help?
Amy :)

You can manipulate the square roots, but this is not really one calculates such limits in practice. The practical way is to insert for x: 5 + epsilon and then expand in powers of epsilon.

In the case of a fraction, like this case, this reduces to L'Hopital's rule: Differentiate the numerator and denominator separately and then insert the limiting value.

If (a) = 0 then:

Lim x-->a f(x)/g(x) =

Lim h-->0 f(a+h)/g(a+h) =

Lim h-->0 [f(a+h) - (a+h)-g(a)]

Divide numerator and denominator by h inside the limit. The limit of a fraction is the ratio of the limits of the denominator and numerators separately, if that ratio exists.

To find the limit of the given function, we can use L'Hopital's rule since we have an indeterminate form of 0/0 when substituting 5 for x.

Here's how you can apply L'Hopital's rule in this case:

1. Start by differentiating the numerator and denominator separately with respect to x.

Numerator: The square root of (9 - x) - 2
Differentiating the numerator, we get: (1/2)*(9 - x)^(-1/2)*(-1) = -1/(2*sqrt(9 - x))

Denominator: x - the square root of (x + 20)
Differentiating the denominator, we get: 1 - (1/2)*(x + 20)^(-1/2)*1 = 1 - 1/(2*sqrt(x + 20))

2. Now substitute the limiting value, which is 5, into the derivative expressions.

Numerator: -1/(2*sqrt(9 - 5)) = -1/4

Denominator: 1 - 1/(2*sqrt(5 + 20)) = 1 - 1/(2*sqrt(25)) = 1 - 1/10 = 9/10

3. Take the ratio of the limits of the numerator and denominator as h approaches 0.

Lim h->0 (-1/4) / (9/10) = -10/36 = -5/18

Therefore, the limit as x approaches 5 of the given function is -5/18.

In summary, you can use L'Hopital's rule to find limits of functions that have an indeterminate form, such as 0/0 or infinity/infinity. By differentiating the numerator and denominator and substituting the limiting value, you can determine the value of the limit.