The problem was to prove
ln(ab) = ln(a) + ln(b)using only our limit definition of ln(x), which, as you probably recall is
xh - 1 ln(x) = limStep 1: Apply the limit definition to ln(ab).h> 0 h
You simply substitute ab for x in the limit equation. You would get
(ab)h - 1 ln(ab) = limh> 0 h
Step 2: Apply the identity for a product taken to a power. You
will apply it to (ab)h.
You should have gotten
ahbh - 1 ln(ab) = limh> 0 h
Step 3: Be clairvoyant. It's hard for anybody to explain where the vision comes from to try a step like this. The more proofs and derivations you do, the more easily the vision comes. The objective here is to end up with something in the numerator that we can factor. Specifically we want the numerator to factor into
(ah - 1)(bh - 1)which multiplies out to
ahbh - ah - bh + 1We can't just go changing the numerator to this. But you can introduce the ah term by adding it and subtracting it. That is the same as adding
You should have gotten
ahbh - ah - bh + 1 + ah + bh - 2 ln(ab) = limh> 0 h
Step 4: Factor what you can factor. We already saw that the first
four terms of this numerator can be factored. Factor them and write the
new equation.
You should have gotten
(ah - 1)(bh - 1) + ah + bh - 2 ln(ab) = limh> 0 h
Step 4: Break the fraction in two. Separate the factored stuff from
the residual terms. Write the new equation, then
You should now have
(ah - 1)(bh - 1) ah + bh - 2 ln(ab) = lim+h> 0 h h
Step 5: Observe that the left hand fraction contains a familiar expression. Can you see it as
ah - 1But in the limit, we know that(bh - 1) h
ah - 1becomes ln(a) and (bh - 1) becomes zero. That makes their product zero in the limit as well. So in the limit, the entire left hand fraction is zero. And that leaves you with what? Rewrite the equation without the stuff that goes to zero in the limit, thenh
You should be left with
ah + bh - 2 ln(ab) = limIf you have a good eye for algebra, you might be able to complete the derivation from this point.h> 0 h
Step 6: One plus one equals two. Express the -2 as two
-1's. Then rearrange the terms and break the expression into
the sum of two familiar looking fractions. When you are done,
You should now have
ah - 1 bh - 1 ln(ab) = limThe rest should be a piece of cake.+h> 0 h h
Step 7: Apply the limit formula to each of the new fractions. Remember that the limit of the sum is the sum of the limits. And what is the limit of each of those fractions according to our original limit formula?
xh - 1 ln(x) = limJust substitute the appropriate variable in for x in each case and you geth> 0 h
ln(ab) = ln(a) + ln(b)which is what we were trying to prove all along.
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