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# Find the limit as x to 0 of the given expression

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

# Find the limit as x goes to 0 of ((1+x)1/x – e) / x

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

# Find the limit as x goes to 0 of (x + e2x)1/x

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,

# Find the limit as x goes to 1 of x1 / (1-x)

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,

# Find the limit as x goes to 0 of (ax – asin x) / x3

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.

# Find the limit as x goes to 0 of (ax – 1) / (bx – 1)

Evaluate the limit for .

First we write and . Then we use the expansion of (p. 287), to obtain expansions for and ,

Therefore, we have

# Find all x satisfying equations given in terms of sinh

Let be the number such that . Find all that satisfy the given equations.

1. .
2. .

1. We are given . From the formula for this means

Then, from the given equation we have

Thus,

So, then we have

Therefore we have

2. There can be no which satisfy the given equation. As in part (a), we use the definition of to obtain the equation,

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.

# Derive some properties of the product of ex with a polynomial

Let

1. Prove that

where denotes the th derivative of .

2. Do part (a) in the case that is a cubic polynomial.
3. Find a similar formula and prove it in the case that is a polynomial of degree .

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 .)

1. Proof. From the formula above we have

But, since is a quadratic polynomial we have

Hence, we have

2. If is a cubic polynomial we may write,

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

3. Claim: Let be a polynomial of degree ,

Let . Then,

Proof. Using Leibniz’s formula again, we have

But for the degree polynomial , we know if and for all . Hence, we have

# Prove some properties of a differentiable function satisfying a given functional equation

Let be a function differentiable everywhere such that

1. Prove that and conjecture and prove a similar formula for .
2. Conjecture and prove a formula for in terms of for all positive integers .
3. Prove that and compute

4. Prove that there exists a constant such that

for all . Find the value of the constant .

1. Proof. We can compute this using the functional equation:

Next, we conjecture

Proof. Again, we compute using the functional equation, and the above formula for ,

2. We conjecture

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

3. Proof. Using the functional equation

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

4. Proof. Since the derivative must exist for all (by hypothesis) we know that the limit

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

# Prove that a function satisfying given properties must be ex

Given a function satisfying the properties:

and

Prove the following:

1. The derivative exists for all .
2. We must have .

This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).

1. Proof. To show that the derivative exists for all we must show that the limit

exists for all . Using the given properties of we can evaluate this limit

Therefore, for all , so the derivative is defined everywhere

2. Proof. From part (a) we know . By Section 6.17, Exercise #39 (linked above) we know that the only functions which satisfy this equation are for all or for some constant (where in the linked exercise). However, since the derivative of exists everywhere, and differentiability implies continuity, we know is continuous everywhere. Hence, . Then,

since , so . Therefore, we must have for some constant . Furthermore, we must have since . Thus,