## 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For the linguists among us, the French name “L’ H**ô**pital” was “L’ H**os**pital” but the ‘os’ has been replaced by the ‘o’ with a circumflex accent. Some now write it “L’ H**o**pital”, especially in Internet URLs.

## 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

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

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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