Evaluate the limit.
First, we write the expression using the definition of the exponential,
Now, considering the expression in the exponent and using the expansion of as (page 287 of Apostol) we have as ,
Therefore, we have
Evaluate the limit.
First, we write the expression using the definition of the exponential,
Now, considering the expression in the exponent and using the expansion of as (page 287 of Apostol) we have as ,
Therefore, we have
Evalue the limit.
First, we have
As we use the expansion for (page 287 of Apostol) to write,
From this we see that as ; and so,
Hence, using the expansion for as (page 287 of Apostol) we have
Therefore, we have
Evaluate the limit.
From the definition of the exponential we have
So, first we use the expansion of as (page 287 of Apostol) to write
Therefore, as we have
Now, since as we can use the expansion (again, page 287) of as to write
Therefore, as we have
So, getting back to the expression we started with,
But, as in the previous exercise (Section 7.11, Exercise #23) we know . Hence,
Evaluate the limit.
First, we write
From this exercise (Section 7.11, Exercise #4) we know that as we have
Therefore, as ,
So, we then have
(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes . I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since as we take the expansion of as ,
Therefore,
Evaluate the limit.
First, we want to get expansions for and as . For we write and use the expansion (page 287 of Apostol) of . This gives us
Next, for , again we write and then use the expansion for we have
Now, we need use the expansion for (again, page 287 of Apostol)
and substitute this into our expansion of ,
(Again, this is the really nice part of little -notation. We had lots of terms in powers of greater than 3, but they all get absorbed into , so we don’t actually have to multiply out and figure out what they all were. We only need to figure out the terms for the powers of up to 3. Of course, the 3 could be any number depending on the situation; we chose 3 in this case because we know that’s what we will want in the limit we are trying to evaluate.)
So, now we have expansions for and (in which most of the terms cancel when we subtract) and we can evaluate the limit.
Evaluate the limit for .
First we write and . Then we use the expansion of (p. 287), to obtain expansions for and ,
Therefore, we have
Let be the number such that . Find all that satisfy the given equations.
Then, from the given equation we have
Thus,
So, then we have
Therefore we have
Next, we use the equation given in the problem to write,
Furthermore, we can obtain an expression for by considering
Putting these expressions for and into our original equation we have
But this implies
which is impossible. Hence, there can be no real satisfying this equation.
Let
where denotes the th derivative of .
For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if then the th derivative is given by
So, in the case at hand we have and so
(Since the th derivative of is still for all and .)
But, since is a quadratic polynomial we have
Hence, we have
Claim: If then
Proof. We follow the exact same procedure as part (a) except now we have the derivatives of given by
Therefore, we now have
Let . Then,
Proof. Using Leibniz’s formula again, we have
But for the degree polynomial , we know if and for all . Hence, we have
Let be a function differentiable everywhere such that
for all . Find the value of the constant .
Next, we conjecture
Proof. Again, we compute using the functional equation, and the above formula for ,
Proof. The proof is by induction. We have already established the cases and (and the case is the trivial ). Assume then that the formula holds for some integer . Then we have
Thus, if the formula holds for , it also holds for . Hence, by induction it holds for all integers
Then, since the derivative exists for all (by hypothesis) we know it must exist in particular at . Using the limit definition of derivative, and the facts that and we have
must exist for all . Using the functional equation for we have
But then, from part (c) we know and from this exercise (Section 6.17, Exercise #38) we know
Therefore, we have
Given a function satisfying the properties:
and
Prove the following:
This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).
exists for all . Using the given properties of we can evaluate this limit
Therefore, for all , so the derivative is defined everywhere
since , so . Therefore, we must have for some constant . Furthermore, we must have since . Thus,