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) = limStep 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, theneq. 6.3b-2 h> 0 h
So in this problem, you must show that
(bn)h - 1 bh - 1 limStep 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,= n limeq. 6.3b-3 h> 0 h h> 0 h
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 limStep 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= n limeq. 6.3b-4 h> 0 h h> 0 h
When you substitute
k h =If you make the substitution all the way through the left-hand side and multiply numerator and denominator by n, you geteq. 6.3b-5 n
n (bk - 1) bn - 1 limWhy, you might be asking, didn't you substitute k/n into the subscript of the lim symbol? Because if= n limeq. 6.3b-6 k> 0 k h> 0 h
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 limThe rest is to prove that the limit on the left is equal to the sum of the limits on the right.= lim+ limeq. 6.3b-8 h> 0 h h> 0 h h> 0 h
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 limStep 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= lim+ limeq. 6.3b-9 h> 0 h h> 0 h h> 0 h
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 limStep 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++--= h> 0 h h h h h ah - 1 bh - 1 lim+ limeq. 6.3b-10 h> 0 h h> 0 h
Here's what you should have gotten:
ahbh - ah - bh + 1 ah - 1 bh - 1 limYou 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.++= h> 0 h h h ah - 1 bh - 1 lim+ limeq. 6.3b-11 h> 0 h h> 0 h
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+ limeq. 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)oreq. 6.3b-14a h> 0 h
ah - 1 lim (bh - 1)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 approacheseq. 6.3b-14b h> 0 h
And once you've eliminated that factored quotient, the remaining terms, right and left, balance. So we've proved it.
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