L’ Hôpital’s Rule to Cure Unhealthy Functions
Calculus is a powerful tool for mathematicians and scientists, but it does have an Achilles’ heel. The function in question must be smooth, continuous and defined at the point or interval being examined. A function with a “zero divided by zero” result is undefined. L’ Hôpital’s Rule cannot break the laws of calculus, but it can bend them… in some examples.
For the linguists among us, the French name “L’ Hôpital” was “L’ Hospital” but the ‘os’ has been replaced by the ‘o’ with a circumflex accent. Some now write it “L’ Hopital”, especially in Internet URLs.
Image of Integral Symbol by Mike DeHaan
L’ Hôpital’s Rule Only Treats Certain Ailments
Previous articles introduced differential calculus and Riemann sums for integral calculus. An earlier article found its limits; more accurately, this article found the value of a function at the limit as the independent variable approached a specific X-value.
One particular problem for any real-number function arises for “division by zero”. For example, let F(x) = ((x + a)*(x – a))/(x – a) = (x^2 – a^2)/(x – a); then F(x) is undefined at x=a. But where does that value go on a graph?
Two L’ Hôpital’s Rules for the Price of One
There are two forms of L’ Hôpital’s Rule, because it applies to two situations where F(x) = f(x)/g(x) has difficulties at some point x=c. One version applies to “zero divided by zero”; the other to “infinity divided by infinity”.
L’ Hôpital’s Rule for Division by Zero
Image of X-Squared over X by Mike DeHaan
Suppose F(x) = f(x)/g(x) but the limit as x approaches c of f(x) = limit[ g(x) ] = zero. If f’(x) and g’(x) exist and the limit of x approaching c of f’(x)/g’(x) exists, then the limit[ f(x)/g(x) ] = limit[ f'(x)/g'(x) ].
Here is L’ Hôpital’s method for the function introduced above: F(x) = (x^2 – a^2)/(x – a). Let f(x) = (x^2 – a^2), so f’(x) = 2*x. Let g(x) = (x – a), so g’(x) = 1.
The limit as ‘x’ approaches ‘a’ of f’(x)/g’(x) = limit[ 2*x/1 ] = 2*a, with absolutely no fuss or bother.
L’ Hôpital’s Rule then states that the limit as ‘x’ approaches ‘a’ of f(x)/g(x) also equals 2*a. So we would expect the graph of F(x) to approach “2*a” as ‘x’ approaches ‘a’.
Note that an algebraic approach for the limits would have the same result for the limit as ‘x’ approaches ‘a’ of F(x) = ((x + a)*(x – a))/(x – a).
Simply cancelling “(x – a)” would leave limit[ F(x) = (x + a)/1 ]. As ‘x’ approaches ‘a’, the limit is (a + a)/1 = 2*a.
L’ Hôpital’s Rule for Division by Infinities
"Image of -1/X^2 over -2/x^3" by Mike DeHaan
Now suppose F(x) = f(x)/g(x) but the limit as x approaches c of f(x) = limit[ g(x) ] = infinity, either positive or negative. If f’(x) and g’(x) exist and the limit of x approaching c of f’(x)/g’(x) exists <<bold>>or approaches positive <<italic>>or<</italic>> infinity<</bold>>, then the limit[ f(x)/g(x) ] = limit[ f'(x)/g'(x) ].
Here is an example for a different function F(x) = (a/x)/( b/(x^2) ). As ‘x’ approaches zero, both numerator and denominator approach infinity.
Setting F(x) = f(x)/g(x) = (a/x)/( b/(x^2) ), we need f’(x) and g’(x).
Remember that for any ‘n’, h(x) = a*x^n has the derivative h’(x) = a*n*x^(n-1).
Therefore when f(x) = a/x = a*x^(-1), we find f’(x) = -a*(x^(-2)) = -a/x^2. For g(x) = b/(x^2) = b*x^(-2), we find g’(x) = -2*b*a^(-3) = -2*b/(a^3).
Then limit as ‘x’ approaches zero of f’(x)/g’(x) = limit[ ( (-a/x^2) ) / ( -2*b/(a^3) ) ]. Again both numerator and denominator are heading for infinity.
Therefore by L’ Hôpital’s Rule, the original F(x) approaches infinity as ‘x’ approaches zero.
Extending L’ Hôpital’s Rule
L’ Hôpital’s Rule is transitive: if the first derivatives f’(x)/g’(x) do not help, perhaps the second derivatives f”(x)/g”(x) will resolve the issue. Or the third, or the fourth…
Limits to L’ Hôpital’s Rule
One nasty problem comes if the values for f’(x)/g’(x) oscillate as ‘x’ approaches the critical value; for example if the sign changes between positive and negative.
In other cases, the limit of f(x)/g(x) may seem bounded even though each approaches infinity, but f’(x)/g’(x) might head for infinity.
The main limitation is that the limit of f(x)/g(x) must be “zero divided by zero” or “infinity divided by infinity”.
The Men Behind L’ Hôpital’s Rule
Although only one person’s title graces “L’ Hôpital’s Rule,” in fact two mathematicians contributed to it.
Guillaume François Antoine, Marquis de l’ Hôpital
Picture of Guillaume, Marquis de l'Hôpital by Bemoeial2
L’Hôpital was born in Paris in 1661. As a son of a noble in a high-ranking military family, he entered the army. He resigned as captain of a cavalry regiment due to severe myopia, but had previously shown his talent for mathematics.
At fifteen years of age, he had already solved several problems developed by Blaise Pascal. Although he had continued to study mathematics while enrolled, he was free to fully pursue mathematics after leaving the army. He met, befriended and studied under Jean Bernoulli, one of the very rare breed of folk who understood the newfangled “calculus” of Newton and Leibniz. Many historians believe that l’Hôpital’s references to Bernoulli’s ideas regarding L’Hôpital’s Rule indicate that Bernoulli actually had the original idea.
He was made a member of the Academy of Sciences of Paris and wrote a number of works explaining and developing calculus. L’Hôpital died in 1704.
Jean Bernoulli, aka John Bernoulli
Picture of Jean Bernoulli by Magnus Manske
Jean Bernoulli, born in 1667, earned a philosophy degree and then a medical license at the University of Basel in Switzerland. A member of a family famed for its mathematicians and scientists, he continued the family tradition by teaching the Marquis de l’Hôpital in Paris. On l’Hôpital’s recommendation, he was hired at Gröningen in the Netherlands. Later he took his deceased brother’s place as a professor of mathematics in Basel.
Bernoulli was seen as instrumental in helping develop L’ Hôpital’s Rule. However, he is perhaps most famous for one incident involving several great mathematicians. He posed a difficult challenge, the “brachistochrone” (“quickest time”) about how quickly a bead would slide down a string not dangled vertically and what shape the string would take. Correct answers came from the Bernoulli brothers, l’ Hôpital, Leibniz and an anonymous respondent. Jean Bernoulli identified that correspondent as Isaac Newton, because “I recognize the lion by his paw.”
The solution to the problem, by the way, is “an upside-down cycloid.” But that might be the topic of another article.
The Value of L’ Hôpital’s Rule
L’ Hôpital’s Rule is a very useful tool, when properly used, to find the limit of a function at a point where the function degenerates to either “zero over zero” or “infinity over infinity”. There are limits to its capabilities, even though its capabilities are derived from limits.
References:
Fox, W. Guillaume-François-Antoine de L’Hôpital. The Catholic Encyclopedia. Vol. 7. New York: Robert Appleton Company. (1910). Accessed Aug. 13, 2011.
Weisstein, E. L’Hospital’s Rule. MathWorld. Accessed Aug. 13, 2011.
Integration by Parts. Harvey Mudd College. Accessed Aug. 13, 2011.
John Bernoulli. McGraw-Hill Higher Education. Accessed Aug. 13, 2011.
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