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The purpose of l'Hôpital's rule is to evaluate a limit which is in an indeterminate form.

It is the case where certain limits do indeed converge onto a value, but direct substitution and the traditional algebraic manipulations fail to produce a solution on account of the indeterminate form.

There are two indeterminate forms in which the rule may be used:

• $\tfrac{0}{0}$
• $\pm\tfrac{\infty}{\infty}$

Basic Concept

Suppose for example we have two functions $f(x),g(x)$ , where $f(a)=g(a)=0$ . And we now wish to evaluate the limit:

$\lim_{x\to a}\frac{f(x)}{g(x)}$

This is a case where we might use the Rule because direct substitution yields an indeterminate form $\tfrac{0}{0}$ , and other algebraic techniques are, for the sake of argument, ineffective.

These two functions are defined to be continuous in the vicinity of $a$ , except possibly at $a$ . Otherwise, these two functions are representative of any arbitrary function.

The rule is one that is necessarily taught after differential calculus. Though limit evaluation is generally taught before differentiation, the Rule requires differentiation to be implemented. Differentiation is an added skill which allows us to evaluate limits that we otherwise were unable to.

If the functions $f(x),g(x)$ are both, independently of one another, differentiable functions then the rule may be implemented. That is, $f'(x),g'(x)$ must both exist, at least in the vicinity of $a$ .

The following conditions must be met:

• The two functions which form the numerator and denominator of the rational function are both, on their own, differentiable.
• That is, $\frac{d}{dx}f(x)$ exists and $\frac{d}{dx}g(x)$ exists.
• Direct substitution must produce the indeterminate forms $\tfrac{0}{0}$ or $\pm\tfrac{\infty}{\infty}$

Upon meeting the criteria, we may implement the Rule.

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\frac{df}{dx}}{\frac{dg}{dx}}$

Notice that differentiation was applied once to each function in both numerator and denominator, not once to the quotient.

The result should be evaluable. If the result is another indeterminate form, the Rule may be applied again; that is, the Rule is infinitely reiterative.

Common Mistakes

• Failure to make sure of the indeterminate form.
• Differentiating the entire quotient, or rational expression, as opposed to differentiating the numerator and denominator function separately.
• Forgetting the rule is reiterative.

Proof of the Rule

For the $\tfrac{0}{0}$ Indeterminate Form

The dilemma which causes our difficulty, thereby necessitating the rule, is the indeterminate form $\tfrac{0}{0}$ . We know, then, by definition, that $f(a)=g(a)=0$ .

First by stating the problem:

$\lim_{x\to a}\frac{f(x)}{g(x)}$

We subtract a value of 0 (zero identity of addition) from both the numerator and denominator. This changes nothing. The zero, however, takes the form of their respective function evaluated at $a$ :

$\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}$

Now we multiply the quotient by 1 (unit identity of multiplication) in the form of a rational expression of equal numerator and denominator. We will use $x-a$ :

$\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}\cdot\frac{x-a}{x-a}$

Which we algebraically rearrange:

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\cdot\frac{x-a}{g(x)-g(a)}$

We may reciprocate the right hand quotient by dividing it instead:

$\lim_{x\to a}\frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}$

According to the limit properties, the limit of a product (or quotient) equals the product (or quotient) of the limits, if they each exist and the denominator isnt 0:

$\lim_{x\to a}\left[\frac{\lim\limits_{x\to a}\dfrac{f(x)-f(a)}{x-a}}{\lim\limits_{x\to a}\dfrac{g(x)-g(a)}{x-a}}\right]$

Notice the additional limit on the left encompassing the whole. We may do this because limits are idempotent operations. We must do this in order to account for the possibility that the denominator may indeed evaluate to zero, a stipulation otherwise not permitted when we split the limit of a quotient into a quotient of limits. As a limit still, we may disregard the zero-denominator possibility.

In other words, by encompassing the whole with another limit we are able to split the limits apart without concern for what the denominator evaluates to. Furthermore, we must still evaluate the limit as the limiting variable $x$ approaches $a$ - without the outermost limit the $a$ is lost to us.

Both of these quotient limits are the limit definitions of the derivative (via secant).

$\lim_{x\to a}\frac{\frac{df}{dx}}{\frac{dg}{dx}}$

For the $\tfrac{\pm\infty}{\pm\infty}$ Indeterminate Form

Due to the fact that the ± sign is nothing more than a multiplicative factor valued at $\pm1$ , it is a constant and a scalar which may be factored out of the limits. Consequently, the rule which applies to the indeterminate form $\tfrac{\infty}{\infty}$ also applies to $\tfrac{-\infty}{\infty}\ ,\ \tfrac{\infty}{-\infty}\ ,\ \tfrac{-\infty}{-\infty}$ .

We may assume now that as $x\to a$ , our functions $f(x),g(x)$ both, separately, approach an infinite magnitude.

$\lim_{x\to a}\frac{f(x)}{g(x)}\to\frac{\infty}{\infty}$

What we may do is perform algebraic manipulations. We shall copy from the following algebraic rule:

$\frac{a}{b}=\frac{\frac{1}{b}}{\frac{1}{a}}$

As it pertains to the limit in question:

$\lim_{x\to a}\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}$

Now, even though the functions $f,g$ approach infinite magnitudes, their reciprocals approach zero from the positive.

$\lim_{x\to a}\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}\to\frac{\frac{1}{\infty}}{\frac{1}{\infty}}=\frac{0^+}{0^+}$

We are now in an indeterminate form $\tfrac{0}{0}$ , and l'Hopitals Rule for $\tfrac{0}{0}$ applies.

$\lim_{x\to a}\frac{\frac{d}{dx}\frac{1}{g(x)}}{\frac{d}{dx}\frac{1}{f(x)}}$

Evaluate the derivatives, applying the chain rule.

$\lim_{x\to a}\frac{-\frac{g'(x)}{g(x)^2}}{-\frac{f'(x)}{f(x)^2}}$

Simplify:

$\lim_{x\to a}\frac{f(x)^2\cdot g'(x)}{g(x)^2\cdot f'(x)}$

We have thus far proved the last equality.

We may apply the various limit properties in order to rearrange the functions. We can end up with:

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$

Dealing with Other Indeterminate Forms

There are five other indeterminate forms involving the algebraic relationships. The l'Hopital's Rule only applies to those two involving division:

• $\tfrac{0}{0}$
• $\tfrac{\infty}{\infty}$

But the other five are:

• $0\cdot\infty$
• $\infty-\infty$
• $0^0$
• $1^\infty$
• $\infty^0$

In order to solve limits such as these requires algebraic manipulations in order to get them into the $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ forms so that l'Hopitals Rule may be applied.

The $0\cdot\infty$ Form

If $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=\infty$ then we may change one function from a multiplicative factor into the division of a reciprocal.

$\lim_{x\to a}f(x)\cdot g(x)=\lim_{x\to a}\frac{g(x)}{\frac{1}{f(x)}}=\lim_{x\to a}\frac{f(x)}{\frac{1}{g(x)}}$

The limit goes from being a $0\cdot\infty$ indeterminate to being a $\tfrac{\infty}{\infty}$ or $\tfrac{0}{0}$ , respectively.