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    148354
  • 3.3 L'H\^{o}pital's rule

    Consider the limit

    eq_b58b2df1.png

    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,

    eq_490736f0.png

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

       N(Sin(10^-20)/10^-20,50)
        0.99999999999999999 
        9999999999999999999 
        99998333333333 
    

    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.

    sin-x-over-x.jpg

    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.

    eq_6710cf49.png

    eq_8db557a4.png

    eq_0f9312db.png

    eq_1ee5a236.png

    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
      1+(-0.16667)d^2+... 
    

    (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'H\^{o}pital's rule (simplest form)

    pital's rule (simplest form)} If u and v are functions with u(a)=0 and v(a)=0, the derivatives v·(a) and v·(a) are defined, and the derivative v·(a)≠ 0, then
    eq_8d1e0dc4.png

    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

    dx} qquad ,

    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 u·, and likewise for dv/dx, so this establishes the result.

    We will generalize L'H\^{o}pital's rule on p. 65.

    By the way, the housetop accent on the “\^{o}” in l'H\^{o}pital 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 11

    As remarked above, the example of \lim{x→0} 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

    eq_827e5ed3.png

    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 12

    ◊ Evaluate

    eq_09b733b6.png

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

    Example 13

    ◊ Evaluate

    eq_8335a941.png

    ◊ 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

    eq_3ed94417.png