Differential and Integral Calculus

More general versions.

If the hypotheses hold only from one side, for example \(\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = 0\text{,}\) then the conclusion still holds on that side: if \(\lim_{x \to a^+} f'(x)/g'(x) = L\) then \(\lim_{x \to a^+} f(x)/g(x) = L\text{.}\) Also, the limit can be taken at \(\pm \infty\) and nothing changes.

Turning other indeterminate forms into \(0/0\).

For each of the other indeterminate forms, there's a trick to turn it into the basic form 0/0.

The case \(0 \cdot \infty\)

Suppose \(\lim_{x \to a} f(x) = 0\) and \(\lim_{x \to a} g(x) = \infty\text{.}\) How can we compute \(\lim_{x \to a} f(x) \cdot g(x)\text{?}\) We know that \(\lim_{x \to a} 1/g(x) = 1/\infty = 0\text{.}\) Therefore, an easy trick is to replace multiplication by \(g\) with division by \(1/g\text{.}\) Letting \(h\) denote \(1/g\text{,}\) we have

\begin{equation*} \lim_{x \to a} f(x) g(x) = \lim_{x \to a} \frac{f(x)}{h(x)} \end{equation*}

which is the correct form for L'Hôpital's Rule.

Example 7.6.

What is \(\lim_{x \to 0^+} x \cot x\text{?}\) Letting \(f(x) = x\) and \(g(x) = \cot x\) we see this has the form \(0 \cdot \infty\text{.}\) Letting \(h(x) = 1/g(x) = \tan x\) we see that

\begin{equation*} \lim_{x \to 0^+} x \cot x = \lim_{x \to 0^+} \frac{x}{\tan x} = \lim_{x \to 0^+} \frac{x}{\sin x} \cdot \cos x \, . \end{equation*}

The limit at 0 of \(x / \sin x\) is~1 and the limit of the continuous function \(\cos x\) is \(\cos (0) = 1\text{,}\) therefore the answer is \(1 \cdot 1 = 1\text{.}\)

The case \(\infty / \infty\)

You could invert both \(f\) and \(g\text{,}\) writing \(\frac{f(x)}{g(x)}\) as \(\frac{1/g(x)}{1/f(x)}\text{.}\) There is a reasonable chance that L'Hôpital's Rule can be applied to this. There is also another version of L'Hôpital's Rule specifically for this case.

Theorem 7.7. L'Hôpital's Rule for \(\infty / \infty\).

If both \(f\) and \(g\) tend to \(\infty\) or \(-\infty\) as \(x \to a\text{,}\) then

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \end{equation*}

whenever the right-hand side is a real number or \(\pm \infty\text{.}\)

Checkpoint 91.

Compute \(\lim_{x \to \infty} \, \frac{x}{e^x}\) .

Answer.

\(0\)

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The cases \(1^\infty\text{,}\) \(0^0\) and \(\infty^0\)

The idea with indeterminate powers is to take the log, compute the limit, then exponentiate. The reason this works is that \(e^x\) is a continous function. Theorem 4.31 says that if \(\lim_{x \to a} h(x) = L\) then \(\lim_{x \to a} e^{h(x)} = e^L\text{.}\)

The way we will use this when evaluating something of the form \(\lim_{x \to a} f(x)^{g(x)}\) is to take logarithms. Algebra tells us \(\ln f(x)^{g(x)} = g(x) \ln f(x)\text{.}\) If we can evaluate \(\lim_{x \to a} g(x) \ln f(x) = L\) then we can exponentiate to get \(\lim_{x \to a} f(x)^{g(x)} = e^L\text{.}\)

Checkpoint 92.

Suppose \(\lim_{x \to a} g(x) = 0\) and \(\lim_{x \to a} f(x)\) is either undefined or a real number other than 0. What do you think can be said about \(\lim_{x \to a} f(x) / g(x)\text{?}\) Proof not needed -- it’s OK to guess on this one.

Example 7.8. continuous compounding.

Suppose you have a million dollars earning a 12\% annual interest rate for one year. You might thing after a year you will have \(1.12\) million dollars. But no, things are better than that. The bank compounds your interest for you. They realize you could have cashed out after half a year with 1.06 million and reinvested for another half year, giving you \(1.1236\) million, which doesn't seem so different but is actually 3600 dollars more. You could play this game more frequently, dividing the year into \(n\) periods and earning \(12\% / n\) interest \(n\) times, so your one million becomes \((1 + 0.12/n)^n\) million.

With computerized trading, you could make the period of time a second, or even a microsecond. Does this enable you to claim an unboundd amount of money after one year? To answer that, let's compute the amount you would get if you compounded {\mathbb{E}m continuously}, namely \(\lim_{n \to \infty} (1 + 0.12/n)^n\text{.}\) Taking logs gives \(\ln (1 + 0.12/n)^n = n \ln (1 + 0.12/n)\text{.}\) Changing to the variable \(x := 1/n\text{,}\)

\begin{align*} \lim_{n \to \infty} n \ln (1 + 0.12/n) & = \lim_{x \to 0} \frac{\ln (1 + 0.12 x)}{x} \\ & = \lim_{x \to \infty} \frac{(d/dx) \ln (1 + 0.12 x)}{(d/dx) x} \mbox{ (L'Hôpital's Rule)} \\ & = \lim_{x \to \infty} \frac{0.12 / (1 + 0.12x)}{1} \mbox{ (use the chain rule)}\\ & = \; 0.12 \; \end{align*}

Therefore, \(\lim_{n \to \infty} (1 + 0.12/n)^n = e^{0.12} \approx 1.12749685\) million dollars. That's better than the $120,000 you earn without compounding, or the $3,600 more than that you earn compounding once, but it's not infinite, it's just another $3896.85 better.

Checkpoint 93.

What is \(\lim_{t \to 0} \, (1+t)^{1/t}\text{?}\) This limit is sometimes used to define the famous constant named after Euler.

Answer.

\(e\)

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Repeated use of L'Hôpital's Rule.

Sometimes when trying to evaluate \(\lim_{x \to a} f(x) / g(x)\) you find that \(\lim_{x \to a} f'(x) / g'(x)\) appears a bit simpler, but you still can't tell what it is. You might try L'Hôpital's Rule twice. If \(f'(x)\) and \(g'(x)\) tend to zero as \(x \to a\) (if they don't, you can probably tell what the limit is), then you can use \(f'\) in place of \(f\) and \(g'\) in place of \(g\) in L'Hôpital's Rule. If you can evaluate the limit of \(f'(x) / g'(x)\) then this must be the limit of \(f(x)/g(x)\text{.}\) You can often do a little better if you simplify \(f'(x) / g'(x)\) to get a new numerator and denominator whose derivatives will be less messy.

Comparisons at infinity.

Discussion.

This is a general rule: the function \(g(x) + h(x)\) will be asymptotic to \(g(x)\) exactly when \(h(x) \ll g(x)\text{.}\) Why? Because \((g(x) + h(x)) / g(x)\) and \(h(x)/g(x)\) differ by precisely 1. It follows that if \(g(x) + h(x) \sim g(x)\) then

or in other words, \(h \ll g\text{.}\) The chain of identies runs backward as well: \(g + h \sim g\) if and only if \(h \ll g\text{.}\)

Another principle is that if \(f \sim g\) and \(h \sim \ell\) then \(f \cdot h \sim g \cdot \ell\text{.}\) This is literally just the product rule for limits. The same is true for nonzero quotients, for the same reason.

These two facts give important techniques for estimating. They allow you to clear away irrelevant terms: in any sum, every term that is much less than one of the others can be eliminated and the result will be asymptotic to what it was before. You can keep going with products and quotients.

Beware though, if \(f \sim g\) and \(h \sim \ell\text{,}\) it does not follow that \(f + h \sim g + \ell\) or \(f - h \sim g - \ell\text{.}\) Why? See the following self-check exercise.

It should be obvious that the relation \(\sim\) is symmetric: \(f \sim g\) if and only if \(g \sim f\text{.}\) Formally,

because one is the reciprocal of the other. On the other hand, the relation \(f \ll g\) is anti-symmetric: it is not possible that both \(f \ll g\) and \(g \ll f\text{.}\)

It is good to have an understanding of the relative sizes of common functions. Here is a summary of some basic facts from this lesson, practice problems and homework problems.

"For sufficiently large \(x\)".

Often when discussing comparisons at infinity we use the term "for sufficiently large \(x\)". That means that something is true for every value of \(x\) greater than some number \(M\) (you don't necessarily know what \(M\) is). For example, is it true that \(f \ll g\) implies \(f < g\text{?}\) No, but it implies \(f(x) < g(x)\) for sufficiently large \(x\text{.}\) Any limit at infinity depends only on what happens for sufficiently large \(x\text{.}\)