Section 8: More Tricks with Derivatives© 1998 by Karl Hahn |
Back in the 1960's, the banks in the United States were regulated by the federal government. One of the regulations placed a ceiling of 6% per annum on interest that they could pay depositors on passbook savings accounts. By the late 1960's economic forces were causing funds available to the banks to be in short supply. They all raised their deposit interest rates to the maximum in order to attract depositors. They also gave away toasters and radios to new depositors.
As money grew even tighter, the banks looked for loopholes in the 6% maximum regulation. And what they found was that there was no regulation on how often they could compound the interest. Traditionally they had been compounding annually. So $100 left on deposit for one year earned $6. Now they started compounding quarterly. So in effect, they were paying 1.5% per quarter. This way, $100 left on deposit for three months earned $1.50. So after three months, your account balance would be $101.50. The next three months would earn 1.5% of $101.50, or $1.52. So six months into the year, your $100 would become $103.02. From six months to nine months, you would earn interest on that amount, and end up with $104.57. And from nine months to the end of the year you would earn interest on that amount and end up with $106.14. So instead of earning just 6% per annum on your money, you were effectively earning 6.14% on your money.
Soon all the banks were offering quarterly compounding. To compete for deposits, some banks began offering monthly compounding. This is equivalent to giving you 0.5% on your money each month. And when you work it out, you end up with an effective per annum rate of 6.17%
Well, from monthly compounding it went to daily compounding. That gives you an effective rate of 6.18%. Finally a few banks offered continuous compounding. "Your money earns interest by the second," the ad copy went. So what is the effective per annum rate of compounding 6% per annum continuously?
First, of course, we must come to terms with what is meant by continuous compounding. If we are going to compound a 6% rate n times per year, then we divide the year into n equal intervals and pay 6/n% at the end of each interval. That is the same as multiplying the account balance at the end of each interval by 1 + 0.06/n. And we are going to do that n times in the course of the year. If you start at the beginning of the year with a balance of $100, then at the end of the year, you would have a balance, P of
P = $100 × (1 + 0.06/n)^{n} eq. 8.1-1So the effective per annum rate, r, in percent, would be
r = 100% × ( (1 + 0.06/n)^{n} - 1) eq. 8.1-2The logical thing to do to find what happens when you compound continuously would be to see what limit the above expressions go to when n grows without limit. The crucial expression whose limit we would like to find is (1 + 0.06/n)^{n}. Let
y = lim n → ∞ |
(1 + 0.06/n)^{n} eq. 8.1-3 |
Now we are stuck with the problem of taking this limit.
From what you learned in the section on exponentials and logs, you know that we can take the natural log of both sides of equation 8.1-3 and get
ln(y) = lim n → ∞ |
n ln(1 + 0.06/n) eq. 8.1-4a |
If you make the substitution of h = 1/n, then equation 8.1-4a becomes
ln(y) = lim h → 0 |
ln(1 + 0.06h) |
Observe that as h approaches zero, 1 + 0.06h approaches 1, and the natural log of that approaches zero. So the numerator of the above expression approaches zero as h approaches zero. And clearly, so does the denominator. This suggests that a limit might exist. But how do you proceed from here to find it?
Finding the limit of equation 8.1-4b is a specific case of finding the limit, in general, of
lim x → c |
f(x) |
when the limits of both f(x) and g(x) approach zero as x approaches c.
As an illustration of this, let's take the example of f(x) = x^{3} - 1 and g(x) = x^{2} - 1. Both of these go to zero as x approaches 1. Try taking your calculator and computing f(0.9999) and g(0.9999). Divide the second into the first. What do you get? Do the same with f(1.0001) and g(1.0001). What do you get this time? And from those clues, what would you guess the limit of this quotient is as x approaches 1? Now look at figure 8-1. It shows a plot of both these functions right around the point, x = 1. A small delta from x = 1 is shown in red as dx. As long as dx is not zero, we can take the quotient of f(1 + dx) / g(1 + dx) without dividing by zero. Notice that, as shown, we have |
f(1 + dx) = df eq. 8.1-6a
and
g(1 + dx) = dg eq. 8.1-6b
Recall from our discussions of the mean value theorem that if you have a function, f(x), that is continuous and whose derivative, f'(x), exists and is continuous, if you know the value of f(c), you can approximate the value of f(c + h) by
f(c) + h f'(c) eq. 8.1-7aAnd the smaller h is, the more accurate the approximation is. Well here you know that f(1) = 0. And instead of h, we have dx. So f(1 + dx) = df can be approximated by
f(1) + dx f'(1) = dx f'(1) eq. 8.1-7bLikewise, since it is also true that g(1) = 0, you can approximate g(1 + dx) = dg by
g(1) + dx g'(1) = dx g'(1) eq. 8.1-7cSo you should be able to approximate the quotient of f(1 + dx) / g(1 + dx) by
dx f'(1) f'(1)Not only that, the closer to zero you make dx the better the approximation. And the limit of the approximation as dx approaches zero will be exactly equal to the quotient of f'(1) over g'(1). In this case you have f'(x) = 3x^{2}, so f'(1) = 3. Likewise you have g'(x) = 2x, so g'(1) = 2. Putting it all together, you get=eq. 8.1-8 dx g'(1) g'(1)
lim x → 1 |
x^{3} - 1 f'(1) |
lim x → c |
f(x) |
lim x → c |
f'(c) |
Observe that in the diagram, dx and df form two legs of a right triangle. The curve, f(x), very nearly forms the hypotenuse of that triangle. And the smaller dx is, the closer the f(x) curve becomes to being the hypotenuse. So we would expect that df can be found simply by multiplying dx times the slope of the curve -- and that slope is f'(c) (where c is the point at which f(x) crosses the x axis).
Of course you can make the identical argument for finding dg from dx and g'(c). So as dx goes to zero, the ratio of df = f(c + dx) to dg = g(c + dx) must approach the ratio of dx f'(c) to dx g'(c). The dx's, of course, cancel.
Now let's return to problem of interest rates being compounded continuously. We had gotten as far as:
ln(y) = lim h → 0 |
ln(1 + 0.06h) |
ln(y) = lim h → 0 |
ln(1 + 0.06h) |
lim h → 0 |
0.06/(1 + 0.06h) |
When you take the limit as h goes to zero of the right-hand expression, you find that 1 + 0.06h goes to 1, so you are left with
ln(y) = 0.06 eq. 8.1-11bThis means that
y = e^{0.06} = 1.061836547... eq. 8.1-11cHow much better is that than compounding daily? This corresponds to an effective rate of 6.1836547%. Compounding daily corresponds an effective rate of 6.183131%. On a $10,000 savings account, compounding continuously yields about a nickel more per year than compounding daily. Of course, over the years, those nickels add up.
Lets try another problem. We'll find
lim x → 1 |
ln^{2}(x) |
Step 1: What do the numerator and denominator go to? Since ln(1) = 0, the numerator is clearly zero when x = 1. And you should be able to convince yourself that the denominator is also zero when x = 1. Since they both go to zero, and since both the numerator and denominator are continuous and have continuous derivatives at x = 1, this quotient looks like a good candidate for applying L'Hopital's Rule.
Step 2: Take the derivatives of the numerator and denominator. You can apply the chain rule to the numerator to find that its derivative is
2 ln(x)And the derivative of the denominator is an easy application of the chain rule. For the derivative of the denominator you get 2(x - 1).x
Step 3: Apply L'Hopital's Rule. The rule says that the limit of the numerator divided by the denominator is the same as the limit of the derivative of the numerator divided by the derivative of the denominator. So you have
lim x → 1 |
ln^{2}(x) |
lim x → 1 |
2 ln(x) |
Step 4: Take the limit of the new quotient. First, cancel common the factor of 2 from the numerator and denomonitator. Trouble is, after you do that, the new quotient still has numerator and denominator of zero when x = 1. What to do? Well we just run through the same steps again.
Step 5: Find the derivatives of the numerator and denominator of the new quotient. We know that the derivative of ln(x) is 1/x. That takes care of the numerator. The denominator, x(x - 1) is the same as x^{2} - x. And the derivative of that is 2x - 1.
Step 6: Apply L'Hopital's Rule for a second time. This gives you
lim x → 1 |
ln(x) |
lim x → 1 |
1 |
Step 7: Take the limit. The numerator of the new new quotient is 1 no matter what you put in for x. If you put x = 1 into the denominator, you get 1 as well. So this limit is equal to 1.
The point of the above example is to illustrate that with L'Hopital's
rule, sometimes once is not enough. If you apply it once and both the
numerator and denominator are still zero, simply apply it again, as we did
here. If numerator and denominator are still both zero, apply yet
again. And keep doing that until you end up with a quotient that has either
numerator, denominator, or both not equal to zero. If what you end
up with has a nonzero denominator, then you can divide out the quotient
and produce a limit. If the denominator ends up as zero but the numerator
ends up nonzero, then the quotient has no limit.
1) Use L'Hopital's Rule to find the limits of the following:
a) lim x → 0 |
e^{x} - 1 |
b) lim x → 0 |
e^{x} - 1 - x |
c) lim x → 0 |
e^{x} - 1 - x - (1/2)x^{2} |
d) lim x → 0 |
e^{x} - 1 - x - (1/2)x^{2} - (1/6)x^{3} |
Can you discern a pattern here? If I had put up an example e), what do you suppose it would have been? How would its limit compare to the limits of the above examples? Click here to view answers.
2) Show how L'Hopital's Rule produces the expected result when applied to
lim x → c |
f(x) - f(c) |
Assume that f(x) is continuous and differentiable over some open interval that includes x = c, and that f'(x) is continuous there as well. Click here to view solution.
3) Find the limit:
lim x → 1 |
arcsin(x) - π/2 |
Click here to view solution.
4) Find the limit of
lim x sin(1/x) x → ∞Click here to view solution.
We already know that if you have a quotient whose numerator and denominator both go to zero in the limit, you need to analyze the problem further in order to determine the limit of the quotient. L'Hopital's Rule is a tool for doing such analysis. But it is also true that if both the numerator and denominator of a quotient grow without limit (that is they go to infinity) then you also need to further analyze the problem in order to determine the limit of the quotient.
It turns out that L'Hopital rules in this case as well.
Here's why. Suppose you have two differentiable functions
f(x) and g(x) , both of which
increase without limit as x approaches c.
They are allowed to increase in either the positive or negative
direction (that is, we are allowing them to go to either plus
or minus infinity as x goes to c, and we will
even allow one to go to plus infinity and the other to go to
minus infinity). We would like to find
lim x → c |
f(x) |
But this is the same as
lim x → c |
1 |
Observe that both the numerator expression, 1/g(x), and the denominator expression, 1/f(x), go to zero as x approaches c. Why? Because both g(x) and f(x) go to infinity (or minus infinity) as x approaches c. And because both numerator and denominator of equation 8.1-15b go to zero, we can apply L'Hopital to it by taking the derivative of both the numerator and denominator.
Since both 1/g(x) and 1/f(x) are composite functions, we shall use the chain rule to find their derivatives. And you find that the derivative of 1/g(x) is -g'(x)/g^{2}(x) and the derivative of 1/f(x) is -f'(x)/f^{2}(x). So we have
lim x → c |
f(x) |
lim x → c |
1 |
lim x → c |
g'(x) − |
or, using a little algebra to simplify this, we have
lim x → c |
f(x) |
lim x → c |
f^{2}(x)g'(x) |
Now if we divide through by f^{2}(x) and multiply through by g^{2}(x), we end up with
lim x → c |
g(x) |
lim x → c |
g'(x) |
And if you take the reciprocal of both sides you get
lim x → c |
f(x) |
lim x → c |
f'(x) |
This is just L'Hopital's Rule all over again.
In other words, we have used L'Hopital's Rule for quotients whose numerator and denominator both go to zero in the limit, and we have shown that it is a necessary consequence that L'Hopital's Rule applies also to quotients whose numerator and denominator both go to infinity (or minus infinity) in the limit (you will recall that we prefaced this whole thing by with the condition that both f(x) and g(x) go to either infinity or minus infinity when x approaches c).
Here is an example. Find the limit of
lim x → ∞ |
e^{x} |
Step 1: Confirm that both numerator and denominator grow without limit. That's pretty easy. We know that e^{x} gets very big very fast as x increases, and there is no limit to how big it can get. Since both numerator and denominator have an e^{x} term in them, clearly both numerator and denominator go to infinity as x goes to infinity.
Step 2: Find the derivatives of numerator and denominator. Since e^{x} is its own derivative, we have that both numerator and denominator have derivatives of e^{x}
Step 3: Apply L'Hopital's Rule. That means that the limit of the quotient of numerator over denominator is equal to the limit of the quotient of the derivative of the numerator over the derivative of the denominator.
lim x → ∞ |
e^{x} |
lim x → ∞ |
e^{x} |
Step 4: Take the limit. Clearly the e^{x} in the numerator of the right-hand expression cancels with the e^{x} in the denominator. So
lim x → ∞ |
e^{x} |
5) You may have heard at one time or another that 0^{0} is undefined. All other numbers taken to the zero power, though, yield unity:
x^{0} = 1 for all x ≠ 0But you also have that zero raised to any positive power is zero.
0^{x} = 0 for all x > 0If you take the limit of the first expression as x approaches zero you'd expect it to go to the same limit as when you take the limit of the second expression as x approaches zero. But this does not happen. The limit of the first is 1 while the limit of the second (as x approaches zero from above) is 0. That is why we have to leave 0^{0} undefined. But just for fun, try using a third limit to see what possibilities there are for 0^{0}. Find the limit of
lim x^{x} x → 0 ^{+}Hint: Recall from your study of logs and exponentials that this is the same as
lim e^{x ln(x)} x → 0 ^{+}so it is useful to find the limit as x approaches zero (from above) of x ln(x). And for this, you can apply L'Hopital's Rule. Click here to view solution.
6) Find the limit of
lim x → ∞ |
A_{0} + A_{1}x |
Where A_{0}, A_{1}, B_{0}, and B_{1} are all constants, and A_{1}B_{1} ≠ 0. Click here to view solution.
7) Suppose that P(x) and Q(x) are both polynomials of the same degree, n. That is,
P(x) = A_{0} + A_{1}x + A_{2}x^{2} + ... + A_{n}x^{n}and
Q(x) = B_{0} + B_{1}x + B_{2}x^{2} + ... + B_{n}x^{n}where the A's and B's are all constants, and B_{n} ≠ 0. Use induction and together with the result you got in problem 6 to show that
lim x → ∞ |
P(x) A_{n} |
for any degree, n, you might choose for both polynomials, P and Q. Click here to view solution.
8) Find the limit of
lim x e^{-x} x → ∞Click here to view solution.
9) Use induction along with the result from problem 8 to show that
lim x^{n} e^{-x} = 0 x → ∞for any natural number, n, no matter how large n is. Click here to view solution.
Limits using Approximate Forms© 1998 by Karl Hahn |
Here's an example of a problem that L'Hopital's Rule does not immediately apply to but you wish it would.
_______ lim √x^{2} + 5x - x eq. 8.1-19 x → ∞L'Hopital's Rule applies to ratios, but this is a difference. The trick here is to use the same principles that we used to develop L'Hopital's Rule, but apply them to this difference. In the development of L'Hopital's Rule we approximated f(x+h), where f(x) = 0, as h f'(x). This is just a special case of an approximation we have used before:
f(x + h) ≈ f(x) + h f'(x)where the symbol, ≈, means "is approximately equal to." Likewise by subtracting f(x) from both sides
f(x + h) - f(x) ≈ h f'(x)The only thing special about L'Hopital's use of this is that he looked at the case where f(x) = 0. We will use the more general case.
We know that the approximation gets better and better as h goes to zero. But suppose that instead of letting h go to zero, you keep h the same and let x go to infinity instead. As x gets big, h is a tinier and tinier fraction of x. And the approximation above might still apply. That is it will apply if a certain limit does indeed exist, namely
lim h f'(x) = x → ∞ |
lim f(x + h) - f(x) eq. 8.1-20 x → ∞ |
If the limit on the left exists, it will be equal to the limit on the right.
I know that this sounds a little complicated, but think of it this way. Suppose you had a car whose maximum speed was 50 meters per second. It never actually reaches that speed. But if you take it onto the highway and floor it, 50 meters per second is the limit of how fast it goes. Every minute you keep it floored, the car's speed gets closer to 50 meters per second, and if you floor it long enough it will get as close to 50 meters per second as you like.
Now think about the distance between where the car is at any moment and where it will be 2 seconds hence. That distance also has a limit. And that limit is 2 seconds times 50 meters per second, which is 100 meters. It only makes sense.
That is all that equation 8.1-20 is saying. With the car, f(t) is the car's position at any time. f'(t) is the car's speed. h is the 2 seconds. Think about this until you understand it.
Now we turn it up a notch. Suppose you have a composite:
f( g(x) + h(x) )Let's suppose that g(x) and h(x) both go to infinity as x goes to infinity. Further suppose that f( g(x) + h(x) ) also goes to infinity as g(x) and h(x) do. But here is the kicker. g(x) goes to infinity faster than h(x). Or in other words, as x gets big, g(x) grows larger than h(x) until it dwarfs it. In equations you can state this condition this way:
lim x → ∞ |
h(x) |
So, if you have all these conditions, then what is the limit of
lim f(g(x) + h(x)) - f(g(x)) eq. 8.1-22 x → ∞You apply the same principle, but in equation 8.1-21 you replace x with g(x) and replace h with h(x).
lim f(g(x) + h(x)) - f(g(x)) = x → ∞ |
lim h(x) f'(g(x)) eq. 8.1-23 x → ∞ |
Here is equation 8.1-19 again:
_______ lim √x^{2} + 5x - x eq. 8.1-19 x → ∞Observe that for positive x
__ x = √x^{2}so equation 8.1-19 is really
_______^{ } __ lim √x^{2} + 5x - √x^{2} eq. 8.1-19a x → ∞Note also that
lim x → ∞ |
5x |
If f() is taking the square root, g(x) = x^{2}, and h(x) = 5x, then this fits the model given in equation 8.1-23. So
1 1 f'(g(x)) ==eq. 8.1-24a 2√x^{2} 2x
lim x → ∞ |
h(x) f'(g(x)) = |
lim x → ∞ |
h(x) 5x 5 |
You could have solved this same problem using L'Hopital's rule, but the path is not as direct (using L'Hopital in this case is doing it the hard way. But I'll do it out that way just to show that you end up with the same answer). You have to do some algebraic look-ahead first -- that is you would have to see that it was prudent apply the difference of squares by multiplying by
_______ √x^{2} + 5x + xwhich, of course, is equal to one and doesn't change anything:√x^{2} + 5x + x
_______ lim √x^{2} + 5x - x = x → ∞ |
lim x → ∞ |
x^{2} + 5x - x^{2} |
lim x → ∞ |
5x |
eq. 8.1-25 |
lim x → ∞ |
5 |
which would be fine if it weren't for that nasty fraction on the left side of the denominator. In order to take this limit you have to find out what the limit of the nasty thing is, so you have to apply L'Hopital's rule to it separately. Suppose the limit of the nasty thing is L. Then
L = lim x → ∞ |
2x + 5 |
lim x → ∞ |
2 |
lim x → ∞ |
______ 2√x^{2} + 5 |
eq. 8.1-27 |
The middle expression in the above equation is the left-hand expression after L'Hopital has been applied. The right-hand expression is a simplification of the middle expression.
What you learn from this exercise is that if L is the limit of equation 8.1-27, then L = 1/L. There are only two real numbers with that property. Either L = 1 or L = -1. Since we know x is positive, you can eliminate the latter possibility and go with L = 1. So substitute 1 in for the nasty radical fraction in equation 8.1-26, and you get
lim x → ∞ |
5 5 |
This is the same answer we got doing it using the approximate-form method, but what a lot more work (not to mention a lot more opportunities to make a mistake). The further advantage of the approximate-form method is that it can do differences that this other method can't.
lim (x^{3} + 5x^{2})^{1/3} - x eq. 8.1-29 x → ∞The difference of squares won't help you whip this one into shape for applying L'Hopital. But in the approximate form method, you have f() as taking something to the 1/3 power. You have g(x) = x^{3} and h(x) = 5x^{2}. Clearly this meets the criterion of g(x) growing faster than h(x). You find that f(g(x)) = (x^{3})^{1/3} = x. When you take f'(g(x)) you get
1 f'(g) =Now you can apply the approximate form:g^{-2/3} eq. 8.1-30a 3 1 ^{ } 1 f'(g(x)) =(x^{3})^{-2/3} =eq. 8.1-30b 3 _{ } 3x^{2}
lim (x^{3} + 5x^{2})^{1/3} - x = x → ∞ |
lim x → ∞ |
5x^{2} x + |
5 |
Move on to The Dance of the Derivatives (Related Rates)
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