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 continously?

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``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)So the effective per annum rate,^{n}eq. 8.1-1

r = 100% × ( (1 + 0.06/n)The 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}- 1) eq. 8.1-2

y = lim (1 + 0.06/n)Now we are stuck with the problem of taking this limit.^{n}eq. 8.1-3 n~~> ¥~~

From what you learned in the section on

ln(y) = lim n ln(1 + 0.06/n) eq. 8.1-4a nIf you make the substitution of~~> ¥~~

ln(1 + 0.06h) ln(y) = limObserve that as~~eq. 8.1-4b h~~~~> 0 h~~

Finding the limit of equation 8.1-4b is a specific case of finding the limit, in general, of

f(x) limwhen the limits of both~~eq. 8.1-5 x~~~~> c g(x)~~

As an illustration of this, let's take the example of
g(x) = x^{2} - 1x 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, |

f(1 + dx) = df eq. 8.1-6a

and

g(1 + dx) = dg eq. 8.1-6b

Recall from our discussions of the
`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)`

f(c) + h f'(c) eq. 8.1-7aAnd the smaller

f(1) + dx f'(1) = dx f'(1) eq. 8.1-7bLikewise, since it is also true that

g(1) + dx g'(1) = dx g'(1) eq. 8.1-7cSo you should be able to approximate the quotient of

dx f'(1) f'(1)Not only that, the closer to zero you make~~=~~~~eq. 8.1-8 dx g'(1) g'(1)~~

x^{3}- 1 f'(1) lim~~=~~~~= 1.5 eq. 8.1-9 x~~~~> 1 x~~^{2}- 1 g'(1)

What we have just done above is applied a powerful trick for taking
limits called L'Hopital's Rule (that's
pronounced "LOW - pea - tal". Sometimes you will see it spelled,
L'Hospital, but the pronounciation is the same). L'Hopital's Rule
says this: If `f(x)` and `g(x)` are continuous and
have derivatives that are continuous (on some open interval that
contains `c`), and both
`f(c) = 0``g(c) = 0`

f(x) f'(c) lim |

Figure 8-1 shows why it works. Because both `f` and `g`
are zero at `x = c``x = c``f'(c)`
and `g'(c)` only. As `x` approaches `c`, the
two functions are rather like the Cheshire Cat in *Alice in Wonderland*.
Remember that when he vanished, just before he disappeared, there
was nothing left of the Cheshire Cat except his grin. Here, just before
the two functions vanish, there is nothing left of them but their grins
as well, and you can determine every tooth their disembodied grins
from knowing only their derivatives.

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)``dg = g(c + dx)``dx f'(c)``dx g'(c)``dx`'s, of course, cancel.

Now let's return to problem of interest rates being compounded continuously. We had gotten as far as:

ln(1 + 0.06h) ln(y) = limTo find this limit, you simply apply L'Hopital's rule to it. Here you have~~eq. 8.1-4b h~~~~> 0 h~~

ln(1 + 0.06h) 0.06/(1 + 0.06h) ln(y) = limWhen you take the limit as~~= lim~~~~eq. 8.1-11a h~~~~> 0 h h~~~~> 0 1~~

ln(y) = 0.06 eq. 8.1-11bThis means that

y = eHow 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 continously yields about a nickel more per year than compounding daily. Of course, over the years, those nickels add up.^{0.06}= 1.061836547... eq. 8.1-11c

Lets try another problem. We'll find

ln^{2}(x) lim~~eq. 8.1-12 x~~~~> 1 (x - 1)~~^{2}

**Step 2: Take the derivatives of the numerator and denominator.**
You can apply the

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~~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

ln^{2}(x) 2 ln(x) lim~~= lim~~~~eq. 8.1-13 x~~~~> 1 (x - 1)~~^{2}x~~> 1 2x(x - 1)~~

**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`

**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)``x ^{2} - x`

**Step 6: Apply L'Hopital's Rule for a second time.**
This gives you

ln(x) 1 lim~~= lim~~~~8.1-14 x~~~~> 1 x(x - 1) x~~~~> 1 x(2x - 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``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) eCan you discern a pattern here? If I had put up an example^{x}- 1 lim~~x~~~~> 0 x b) e~~^{x}- 1 - x lim~~x~~~~> 0 x~~^{2}c) e^{x}- 1 - x - (1/2)x^{2}lim~~x~~~~> 0 x~~^{3}d) e^{x}- 1 - x - (1/2)x^{2}- (1/6)x^{3}lim~~x~~~~> 0 x~~^{4}

**2)**
Show how L'Hopital's Rule produces the expected result when applied to

f(x) - f(c) limAssume that~~x~~~~> c (x - c)~~

**3)**
Find the limit:

arcsin(x) - p/2 lim~~x~~~~> 1 Ö1 - x~~^{2}

**4)** Find the limit of

lim x sin(1/x) x~~> ¥~~

We already know that if you have a quotient whose numerator and denominator both go to zero in the limit, you need to analyse 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 analyse 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
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

f(x) limBut this is the same as~~eq. 8.1-15a x~~~~> c g(x)~~

1Observe that both the numerator expression,~~g(x) lim~~~~eq. 8.1-15b x~~~~> c 1~~~~f(x)~~

Since both `1/g(x)` and `1/f(x)` are composite functions,
we shall use the
`1/g(x)` is `-g'(x)/g ^{2}(x)` and the derivative
of

1 g'(x)or, using a little algebra to simplify this, we have~~-~~~~f(x) g(x) g~~^{2}(x) lim~~= lim~~~~= lim~~~~eq. 8.1-15c x~~~~> c g(x) x~~~~> c 1 x~~~~> c f'(x)~~~~-~~~~f(x) f~~^{2}(x)

f(x) fNow if we divide through by^{2}(x)g'(x) lim~~= lim~~~~eq. 8.1-15d x~~~~> c g(x) x~~~~> c g~~^{2}(x)f'(x)

g(x) g'(x) limAnd if you take the reciprocal of both sides you get~~= lim~~~~eq. 8.1-16 x~~~~> c f(x) x~~~~> c f'(x)~~

f(x) f'(x) lim~~= lim~~~~eq. 8.1-17 x~~~~> c g(x) x~~~~> c g'(x)~~

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

e^{x}lim~~eq. 8.1-18a x~~~~> ¥ e~~^{x}+ 1

**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

**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.

e^{x}e^{x}lim~~= lim~~~~eq. 8.1-18b x~~~~> ¥ e~~^{x}+ 1 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}lim~~= 1 eq. 8.1-18c x~~~~> ¥ e~~^{x}+ 1

**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:

xBut you also have that zero raised to any positive power is zero.^{0}= 1 for all x ¹ 0

0If you take the limit of the first expression as^{x}= 0 for all x > 0

lim x^{x}x~~> 0~~^{+}

lim eso it is useful to find the limit as^{x ln(x)}x~~> 0~~^{+}

**6)**
Find the limit of

AWhere_{0}+ A_{1}x lim~~x~~~~> ¥ B~~_{0}+ B_{1}x

**7)**
Suppose that `P(x)` and `Q(x)` are both polynomials
of the same degree, `n`. That is,

P(x) = Aand_{0}+ A_{1}x + A_{2}x^{2}+ ... + A_{n}x^{n}

Q(x) = Bwhere the_{0}+ B_{1}x + B_{2}x^{2}+ ... + B_{n}x^{n}

P(x) Afor any degree,_{n}lim~~=~~~~x~~~~> ¥ Q(x) B~~_{n}

**8)**
Find the limit of

lim x e^{-x}x~~> ¥~~

**9)**
Use induction along with the result from problem 8 to show that

lim xfor any natural number,^{n}e^{-x}= 0 x~~> ¥~~

Here's an example of a problem that L'Hopital's Rule does not immediately apply to but you wish it would.

_______ lim ÖxL'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^{2}+ 5x - x eq. 8.1-19 x~~> ¥~~

f(x + h) » f(x) + h f'(x)where the symbol,

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

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) = lim f(x + h) - f(x) eq. 8.1-20 xIf the limit on the left exists, it will be equal to the limit on the right.~~> ¥ x~~~~> ¥~~

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

h(x) limSo, if you have all these conditions, then what is the limit of~~= 0 eq. 8.1-21 x~~~~> ¥ g(x)~~

lim f(g(x) + h(x)) - f(g(x)) eq. 8.1-22 xYou apply the same principle, but in equation 8.1-21 you replace~~> ¥~~

lim f(g(x) + h(x)) - f(g(x)) = lim h(x) f'(g(x)) eq. 8.1-23 xHere is equation 8.1-19 again:~~> ¥ x~~~~> ¥~~

_______ lim ÖxObserve that for positive^{2}+ 5x - x eq. 8.1-19 x~~> ¥~~

__ x = Öxso equation 8.1-19 is really^{2}

_______Note also that^{ }__ lim Öx^{2}+ 5x - Öx^{2}eq. 8.1-19a x~~> ¥~~

5x limIf~~= 0 eq. 8.1-19c x~~~~> ¥ x~~^{2}

1 1 f'(g(x)) =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~~=~~~~eq. 8.1-24a 2Öx~~^{2}2x h(x) 5x 5 lim h(x) f'(g(x)) = lim~~=~~~~=~~~~eq. 8.1-24b x~~~~> ¥ x~~~~> ¥~~_{ }2x 2x 2

_______ Öx2 + 5x + xwhich, of course, is equal to one and doesn't change anything:~~Öx2 + 5x + x~~

_______ xYou can apply L'Hopital's rule by taking the derivatives of the numerator and denominator to this and you get^{2}+ 5x - x^{2}5x lim Ö x^{2}+ 5x - x = lim~~= lim~~~~x~~~~> ¥ x~~~~> ¥ Öx~~^{2}+ 5x + x x~~> ¥ Öx~~^{2}+ 5x + x eq. 8.1-25

5 limwhich 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~~eq. 8.1-26 x~~~~> ¥ 2x + 5~~~~+ 1 2Öx~~^{2}+ 5x

______ 2x + 5 2 2ÖxThe 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.^{2}+ 5 L = lim~~= lim~~~~= lim~~~~x~~~~> ¥ 2Öx~~^{2}+ 5x x~~> ¥ 4x + 10 x~~~~> ¥ 2x + 5~~~~2Öx~~^{2}+ 5 eq. 8.1-27

What you learn from this exercise is that if `L` is the limit
of equation 8.1-27, then ` L = 1/L`` L = 1 `` L = -1``x` is positive, you can eliminate the latter possibility
and go with ` L = 1``1` in
for the nasty radical fraction in equation 8.1-26, and you get

5 5 limThis 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.~~=~~~~eq. 8.1-28 x~~~~> ¥ 1 + 1 2~~

lim (xThe difference of squares won't help you whip this one into shape for applying L'Hopital. But in the approximate form method, you have^{3}+ 5x^{2})^{1/3}- x eq. 8.1-29 x~~> ¥~~

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}

^{ }5x^{2}5 lim (x^{3}+ 5x^{2})^{1/3}- x = lim x +~~- x~~^{ }=~~eq. 8.1-31 x~~~~> ¥~~_{ }x~~> ¥ 3x~~^{2}3

Move on to The Dance of the Derivatives (Related Rates)

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