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# Prove that (1+x)c = 1 + cx + o(x) as x goes to 0

Prove that

Using this fact compute

Proof. We use the definition and continuity of the exponential to evaluate the limit,

Then, we know (Examples on page 287 of Apostol, taking ) that

Therefore,

Finally we use the expansion, , and Theorem 7.8 (page 288 of Apostol on the algebra of the -symbols) to conclude,

Now, to use this to evaluate the requested limit we make the substitution . Then as and we have

# Compute derivatives of a function given a limit equation that it satisfies

Consider a function that satisfies the limit relation

and has a continuous third derivative everywhere. Determine the values

We do some simplification to the expression first.

Now, we apply the hint (that if then ),

So as we have

But then since has three derivatives at 0 we know its Taylor expansion at 0 is unique and is given by (Theorem 7.1, page 274 of Apostol)

Hence, equating the coefficients of like powers of we have

Next, to compute the limit

we write

Then, using the Taylor expansion of (page 287 of Apostol) we know as we have

Therefore we have

# Prove that 1 / (1+g(x)) = 1- g(x) + g2(x) + o(g2(x))

1. Prove that

where as .

2. Using part (a) prove

Replacing by we have,

By the definition of we have

Therefore,

Hence,

2. Proof. First, we use the expansion of ,

where . By part (a) we then have as ,

Next, since we multiply this expansion for by the expansion (page 287 of Apostol) for as ,

# Prove or disprove given statements for functions such that f(x) = o(g(x))

Let and be functions, both differentiable in a neighborhood of 0, with and such that

Prove or disprove each the following statements.

1. as .
2. as .

1. True.
Proof. Since as we know by the definition of that

Thus, for every there exists a such that

So, for we have

The final line follows since by hypothesis. Therefore,

Hence,

By definition, we then have

2. False.
Consider for and for . Then, for ,

For we have .

Next,

Since we have as . However, since

does not exist.

# Find values for a constant such that a given limit will exist

Consider the limit expression

Find the value of the constant such that this limit will be finite. Find the value of the limit in this case.

We have

In order for this limit to exist we must have ; hence . The limit is then

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

Evaluate the limit.

We use this exercise (Section 7.11, Exercise #4) to evaluate,

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

Evaluate the limit.

We have

# Find the limit as x goes to 0 of (arcsin x / x)1/x2

Evaluate the limit.

First, we rewrite the expression using the definition of the exponential:

Next, we need to get a series expansion for as . The most straightforward way is to take the first few derivatives (we’ll only need the term).

Therefore, we have as

Hence, as

Since this is going to 1 as we may apply this exercise (Section 7.11, Exercise #4) to conclude

Now, we have the expansion as

Therefore,

# 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