L’Hopital’s Rule is a Hospital to Cure Ailing Functions

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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For the linguists among us, the French name “L’ Hôpital” was “L’ Hospital” but the ‘os’ has been replaced by the ‘o’ with a circumflex accent.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

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

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

“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.

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