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3.3. L'Hôpital's Rule

Consider the limit

\[\lim_{x\rightarrow 0} \frac{\sin x}{x} .\]

Plugging in doesn't work, because we get 0/0. Division by zero is undefined, both in the real number system and in the hyperreals. A nonzero number divided by a small number gives a big number; a nonzero number divided by a very small number gives a very big number; and a nonzero number divided by an infinitesimal number gives an infinite number. On the other hand, dividing zero by zero means looking for a solution to the equation 0=0q, where q is the result of the division. But any q is a solution of this equation, so even speaking casually, it's not correct to say that 0/0 is infinite; it's not infinite, it's anything we like.

Since plugging in zero didn't work, let's try estimating the limit by plugging in a number for x that's small, but not zero. On a calculator,

\[\frac{\sin 0.00001}{0.00001} = 0.999999999983333 .\]

It looks like the limit is 1. We can confirm our conjecture to higher precision using Yacas's ability to do high-precision arithmetic:


It's looking pretty one-ish. This is the idea of the Weierstrass definition of a limit: it seems like we can get an answer as close to 1 as we like, if we're willing to make x as close to 0 as necessary. The graph helps to make this plausible.

Figure G. The graph of sin x/x.

The general idea here is that for small values of x, the small-angle approximation sin xx obtains, and as x gets smaller and smaller, the approximation gets better and better, so sin x/x gets closer and closer to 1.

But we still haven't proved rigorously that the limit is exactly 1. Let's try using the definition of the limit in terms of infinitesimals.

\begin{align} \lim_{x\rightarrow 0} \frac{\sin x}{x} &= st \left[ \frac{sin (0+dx)}{0+dx} \right] \\ &= st \left[ \frac{dx+\ldots}{dx} \right] \end{align}

where we've used the identity \(sin(p+q) = \sin p \cos q + \sin q \cos p \). So

\begin{align} \lim_{x\rightarrow 0} \frac{\sin x}{x} &= st \left[1 + \frac{\ldots}{dx} \right] \\ &= 1. \end{align}

In fact, this limit is the same one we would use if we were evaluating the derivative of the sine function, applying the definition of the derivative as a limit.

We can check our work using Inf:

   : (sin d)/d

(The ... is where I've snipped trailing terms from the output.)

Our example involving the limit of sin x/x is a special case of the following rule for calculating limits involving 0/0:

L'Hôpital's Rule (simplest form)

If u and v are functions with u(a)=0 and v(a)=0, the derivatives \(\dot{v}(a)\) and \(\dot{v}(a)\) are defined, and the derivative \(\dot{v}(a)\ne 0\), then

\begin{align} \lim_{x\rightarrow a} \frac{u}{v} = \frac{\dot{u}(a)}{\dot{v}(a)} . \end{align}

Proof: Since u(a)=0, and the derivative du/dx is defined at a, u(a+dx)=du is infinitesimal, and likewise for v. By the definition of the limit, the limit is the standard part of

\[\frac{u}{v} = \frac{du}{dv} = \frac{du/dx}{dv/dx} ,\]

where by assumption the numerator and denominator are both defined (and finite, because the derivative is defined in terms of the standard part). The standard part of a quotient like p/q equals the quotient of the standard parts, provided that both p and q are finite (which we've established), and q ≠ 0 (which is true by assumption). But the standard part of du/dx is the definition of the derivative \(\dot{u}\), and likewise for dv/dx, so this establishes the result.

We will generalize L'Hôpital's rule on p. 65.

By the way, the housetop accent on the “ô” in l'Hôpital means that in Old French it used to be spelled and pronounced “l'Hospital,” but the “s” later became silent, so they stopped writing it. So yes, it is the same word as “hospital.”

Example 42

As remarked above, the example of \(\lim{x\rightarrow0} \sin x/x\) is in some sense circular, since the limit is equivalent to the definition of the derivative of the sine function, so we already need to know the limit in order to evaluate the limit! As an example that isn't circular, let's evaluate

\[ \lim_{x\rightarrow 0} \frac{\sin x}{x+x^3} \]

The derivative of the top is cos x, and the derivative of the bottom is 1. Evaluating these at x=0 gives 1 and 1, so the answer is 1/1=1.

Example 43

◊ Evaluate

\[ \lim_{x\rightarrow 0} \frac{e^x-1}{x} \]

◊ Taking the derivatives of the top and bottom, we find ex/1, which equals 1 when evaluated at x=0.

Example 44

◊ Evaluate

\[ \lim_{x\rightarrow 1} \frac{x-1}{x^2-2x+1} \]

◊ Plugging in x=1 fails, because both the top and the bottom are zero. Taking the derivatives of the top and bottom, we find 1/(2x-2), which blows up to infinity when x=1. To symbolize the fact that the limit is undefined, and undefined because it blows up to infinity, we write

\[ \lim_{x\rightarrow 1} \frac{x-1}{x^2-2x+1} = \infty \]