[FR] Indeterminate Forms (cont.) and l'Hospital's Rule

So a while back, I covered a lesson on Indeterminate Forms. We discussed two common indeterminate quotients (0/0 and ∞/∞) and their troubles when determining limits. Is there a solution...? Yes. The solution is called l'Hospital's Rule.

Find the limit of [x - 49] / [(√x) - 7] as x ---> 49, where √ represents the square root function.


Normally, by the Direct Substitution Property: [49 - 49] / [√(49) - 7] = 0 / [7 - 7] = 0/0

Notice the Direct Substitution Property fails. It gives no means to solve this limit in this form. Algebraic manipulation (such as multiplying by the a / a form of the denominator's conjugate) or attempting to rewrite the equation might work in some cases. However this is a situation where such methods also do not work. Not to mention, they are time consuming and prone to arithmetic errors.

However, l'Hospital's rule can be used to solve this limit.

L'Hospital's Rule:
Suppose the functions f and g are differentiable and g '(x) ≠ 0 on an open interval I that contains a (except possibly at a). Supose that f(x) = 0, x ---> a and g(x) = 0, x ---> a or that f(x) = ± ∞, x ---> a and g(x) = ± ∞, x ---> a (i.e. we have an indeterminate form of type 0 / 0 or ∞ / ∞). Then f(x) / g(x) = f '(x) / g '(x), x ---> a.

This rule states that the quotient of a limit that results in an indeterminate form can be solved by taking the derivatives of the numerator and the denominator and evaluating the limit at x.

Caution: The limit as x ---> a of f(x) / g(x) must be an indeterminate form before applying l'Hospital's Rule.
Caution: L'Hospital's Rule takes f ' and g ' independently. You need not to use the Quotient Rule when formulating the limit.
Caution: The indeterminate form must be a quotient of 0 / 0 or ∞ / ∞

If we apply l'Hospital's Rule on the example:

f (x) = x - 49 ---> f '(x) = 1 - 0 = 1
g(x) = (√x) - 7 = [x ^ (1/2)] - 7 ---> g '(x) = (1/2) x ^ (-1/2) + 0 = 1/(2√x)

Thus: the limit of [x - 49] / [(√x) - 7] as x ---> 49 becomes the limit of 1/(1/2√x) as x ---> 49

Algebraic maniuplation gives us: 2√x as x ---> 49
                                                2√49 = 2 ∙ 7 = 14

How does this work? Well...needless to say, there is a long extensive proof that shows how this property is true. Typing it out would be an extremely cumbersome process, so I leave it up to you if you want to look more into l'Hospital's Rule.

One cool property of l'Hospital's rule is that so long as an indeterminate quotient results as the limit, the process can be repeated multiple times.