Let's take the second identity first -- the one that says
ln(bn) = n ln(b) eq. 6.3b-1since it is a shorter derivation. We must derive this identity starting only with the limit definition of the ln function:
bh - 1
ln(b) = lim eq. 6.3b-2
h > 0 h
Step 1: Set the problem up in terms of the limit definition. That
is, wherever you see "ln" in equation 6.3b-1, rewrite it using the
limit definition given in equation 6.3b-2. See if you can do this step,
then So in this problem, you must show that
(bn)h - 1 bh - 1
lim = n lim eq. 6.3b-3
h > 0 h h > 0 h
Step 2: Apply one of the rules of exponentials. The left-hand
side of equation 6.3b-3 contains a power to a power. We have an identity
for that from section 6.1. Apply that identity and use it to rewrite
equation 6.3b-3. When you are done,
Remember that to find an exponential raised to power, you can simply raise
the base to the product of the powers. That is
bnh - 1 bh - 1
lim = n lim eq. 6.3b-4
h > 0 h h > 0 h
Step 3: Simplify the left side by making a substitution.
As you will see as your calculus studies progress, you have to attack
many problems by substituting a variable with an expression or vice versa.
And that is the approach we will take here. To the beginner, it is not
always easy to see how to make the correct choice for a substitution,
but you get an eye for it the more you practice. In this case it's the
exponent on the left that we need to attack. So substitute
When you substitute
k
h = eq. 6.3b-5
n
If you make the substitution all the way through the left-hand side and
multiply numerator and denominator by n, you get
n (bk - 1) bn - 1
lim = n lim eq. 6.3b-6
k > 0 k h > 0 h
Why, you might be asking, didn't you substitute k/n into the
subscript of the lim symbol? Because if
Step 4: Factor the n out of the limit. We know that the limit of the product is the product of the limits. And the limit of n as h or k goes to zero is simply n. So on the left-hand side of the equals, you can move the n in the numerator to the left of the lim symbol. The left and right hand sides are the same now except that one uses k going to zero and the other uses h going to zero. But does it matter what you call the thing that goes to zero? A rose by any other name would smell as sweet, and a variable that goes to zero by any other name still goes to zero. So the limits are the same. And that completes the derivation.
The other coached exercise was to demonstrate that
ln(ab) = ln(a) + ln(b) eq. 6.3b-7using only the limit definition of the ln function.
Step 1: Rewrite equation 6.3b-7 using the limit definition of
ln. Everywhere you see ln replace it with
the limit definition. Do this, then
This step is just a substitution exercise. You should have gotten
(ab)h - 1 ah - 1 bh - 1
lim = lim + lim eq. 6.3b-8
h > 0 h h > 0 h h > 0 h
The rest is to prove that the limit on the left is equal to the sum of the
limits on the right.
Step 2: Apply a common identity for the exponential of a product.
Remember, you derived that identity in one of the exercises for section 6.1.
When you've done it,
The identity that applies here is
ahbh - 1 ah - 1 bh - 1
lim = lim + lim eq. 6.3b-9
h > 0 h h > 0 h h > 0 h
Step 3: Extract where you're going to from the left-hand side. This
is the not-so-obvious step. The right-hand side of 6.3b-9 is where we want
to go with this. So we are going to add the right-hand side minus itself
to the left-hand side. The net effect is that we are adding zero to the
left (because anything minus itself is always zero), and you can add zero
to anything and it stays the same. And since the sum of the limits is
the limit of the sum, you don't have to repeat the lim notation
five times on the left-hand side of the equals. Once will do just nicely.
Take a stab at this step, then It's starting to look nasty because the equation is getting long, but just be patient and take it one term at a time. You should have gotten
ahbh - 1 ah - 1 bh - 1 ah - 1 bh - 1
lim + + - - =
h > 0 h h h h h
ah - 1 bh - 1
lim + lim eq. 6.3b-10
h > 0 h h > 0 h
Step 4: Rearrange the terms on the left of the equal. We're in
luck because everything is already over the same denominator of h.
That makes the terms easy to add and subtract. Take the negative quotients
and combine them with the first quotient (that is the one with the product
in the numerator). Take care gathering and summing the constant terms
(that is the 1's and -1's). When you've done that
Here's what you should have gotten:
ahbh - ah - bh + 1 ah - 1 bh - 1
lim + + =
h > 0 h h h
ah - 1 bh - 1
lim + lim eq. 6.3b-11
h > 0 h h > 0 h
You can see that the last two quotients on the left of the equals are
looking a lot like what we have to the right of the equals. All we have
to do is prove that that first quotient (the one with all those terms
in the numerator) is actually equal to zero, and then we're very close
to being done.
Step 5: Factor that nasty-looking numerator. It is the product of
two binomials. See if you can find them. Then
The factorization is
ahbh - ah - bh + 1 = (ah - 1)(bh - 1) eq. 6.3b-12So now you have
(ah - 1)(bh - 1) ah - 1 bh - 1
lim + + =
h > 0 h h h
ah - 1 bh - 1
lim + lim eq. 6.3b-13
h > 0 h h > 0 h
Step 6: Convert the sum of quotients into a sum of limits, then
take the limit of the one you factored. To take the limit of
the quotient whose numerator you factored, remember that the limit
of the product is the product of the limits. Your aim is to show
that the limit of the quotient that has the factored numerator is
zero. To show that the limit of a product is zero, you have to
show that both factors have limits and that at least one of those
limits is zero. When you've shown that, you're done.
Then you can
That first quotient can be made to look like either
bh - 1
lim (ah - 1) eq. 6.3b-14a
h > 0 h
or
ah - 1
lim (bh - 1) eq. 6.3b-14b
h > 0 h
Either way, when you take the limit, the quotient part becomes a natural
log (either of b or of a respectively). The other part
of each is zero. Why? Because every positive number raised to
a power approaches 1 as the exponent goes to zero. So in
either case, that term approaches And once you've eliminated that factored quotient, the remaining terms, right and left, balance. So we've proved it.
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