~L'HOSPITAL'S RULE~

~This was originally presented by Marquis De L'Hospital in 1696.  This rule is used for finding special kinds of limits. If you continue to calculus II, it will be covered in depth. I will allow you to use it in my course, but make sure you use it properly.

~A limit, in informal terms, is a specific value that a function gets very close to as the x-values approach a certain value or if x increases (decreases) without bound. If there isn't a limit, then the functional values will either approach positive or negative infinity or possibly get close to more than one value. For a limit to exist, the functional values must cluster around one and only one value (level).

~ L'Hospital's Rule can be used for all types of limits (one-sided & for x approaching positive or negative infinity). However, there is one requirement of each and every one of them. Initially, they must give a 0/0 or ∞/∞ result.  Only then, does that limit meet the hypothesis of L'Hospital's Rule.

~It is used to get limits of the basic indeterminate forms 0/0 or ∞/∞ and others that can be placed into those two basic forms (0⁰, 1^∞, ∞-∞, ∞⁰, 0^∞). The latter forms use log and other algebraic properties to convert them to the two basic forms.

~Many students love the rule so much that they use it when it should not be applied...a big no-no. So, confronted with a 0/0 or ∞/∞ limit, how does the rule work?

~Informally, L'Hospital's Rule states that, if a limit does indeed come out that way, you will get an answer by taking the limit of the ratio of the derivatives of the functions in the top & bottom (limit of the derivative of the numerator divided by the derivative of the denominator). Be careful here, this is not the quotient rule, which is more involved. Here, we simply take the derivative of the top & take the derivative of the bottom, then take the limit again. The result of this limit will be the same as the result of the original limit.

~If this new limit of the derivative over the derivative fails to exist, then the original limit will fail to exist also.

~If this new limit comes out to 0/0 or ∞/∞ again, you can use L'Hospital's Rule again on this one (now you are working with 2nd derivative ratios). As long as the result is 0/0 or ∞/∞, you may apply L'Hospital's Rule. It's not unusual to apply it several times in succession until a result appears or the limit fails to exist.

~Limits that come out to the more advanced indeterminate forms, taking the natural log of both sides & using a few properties can place these into one of the two basic forms, 0/0 or ∞/∞, so you can use the rule. I emphasize again, you must not apply L'Hospital's Rule unless you have one of the two basic forms.

~Let's do some examples

~Ex:  It took us a while to prove that  lim  (sinx)/x = 1 as x→0.
        With L'Hospital's Rule, it can be done in seconds. First, make
        sure it fits one of the two basic forms.  Yes, 0/0. So, this limit will
        have the same answer as lim  (cosx)/1  (which is the derivative of
        the top over the derivative of the bottom) , which is 1. Nice!

~Ex: Take lim   (7x³ + 5x -9) / (3x³ - 2x +3) as x →∞.
        It meets the hypothesis of L'Hosptal's Rule since it comes out to
        ∞/∞.  So, take the lim  (21x² +5)/(9x² -2). This also comes out
        to ∞/∞, so apply the rule again, take the lim  (42x)/(18x) which
        comes out to 7/3, the answer. Nice!. We had short cuts for this one
        without the use of L'Hospital's Rule by comparing the degrees of
        top & bottom, remember?

~Ex: Find lim  x/lnx as x →∞.  We can use the Rule since it
        comes out to ∞/∞.  So, we get, lim  1/(1/x) = lim (x) or ∞. (no
        limit). Therefore, the original limit fails to exist also.

~Note: We can see that L'Hospital's Rule gives us a way of comparing
            the relative "power" or "strength" of expressions by forming their
            ratio & then letting them increase. In the above example, x  will
            increase much faster than lnx, so, it is a more "powerful" term.
            Therefore, the ratio of these two expressions will just get larger
            and no limit will exist.

~Note: I view L'Hospital's Rule as a "power tool" for getting answers to many "weird" limits.

~Note: A word of caution when evaluating limits. Students often seem
            to look for a complicated method before they look for a very
            simple one. Many limits are well behaved & can be found quite
            easily by just substituting into the expression, so, don't over look
            a simple evaluation before you start thinking of some advance
            technique of evaluation.

Proof: To prove: If lim f(x)/g(x)=f(a)/g(a) as x --> a = 0/0 or ∞/∞, Then lim f(x)/g(x) = lim f'(a)/g'(a) as x --> a
Consider lim f(x)/g(x) as x --> a . Let f(a) = g(a) = 0. Then lim f(x)/g(x) = lim [f(x)-f(a)]/[g(x)-g(a)] as x --> a. Now, divide the
numerator and denominator in the limit by (x-a). Since we do not let x=a, this is not zero.
We have, lim f(x)/g(x) = lim {[f(x)-f(a)]/(x-a)}/{[g(x)-g(a)]/(x-a)} as x --> a. Since the limit of a quotient = the quotient of the limits,
we have, lim f(x)/g(x) = lim {[f(x)-f(a)]/(x-a)}/ lim {[g(x)-g(a)]/(x-a)} as x --> a. Both limits define the derivative at x = a for f and g.
Therefore, lim f(x)/g(x) = lim f'(a)/g'(a) as x --> a. For the ∞/∞ case, use the reciprocal t = 1/x. So, this case can be equivalently transformed
into the 0/0 case.