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# 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.

# Prove that sin (1/x) has no limit as x approaches 0

Define the function:

Prove there is no such that

Proof. We prove this by contradiction. Suppose there does exist some such that . From the definition of limit this means that for all there exists a such that

First, we claim that for any such we must have . This must be the case since if then , so we may choose such that . But then, since for all , we have

(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality .) This contradicts our choice of , so must be less than or equal to 1.
Next, suppose . Then choose . To obtain our contradiction we must show that there is no such that

By the Archimedean property of the real numbers we know that for any , there exists a positive integer with such that . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being .) But,

Then, from the definition of and since and we have,

But then,

Furthermore,

This contradicts that if then for every we have for all . In other words, we found an greater than 0 (in particular, we found ) such that no matter how small we choose there exists an such that is smaller than , but is bigger than . This exactly contradicts the definition of . Hence, there can be no such number

This proof shows that there is now way to define so that is continuous at 0. We see this since for to be continuous at 0 we must have . But if we try to define for any , we will fail to achieve continuity since the proof shows for any we might want to choose.

# Give an alternate proof of the continuity of the sine and cosine functions

1. Use the inequality

to prove that the sine function is continuous at 0.

2. Recall the trig identity,

Use this and part (a) to prove that the cosine function is continuous at 0.

3. Use the formulas for sine and cosine of a sum to prove that the sine and cosine functions are continuous for all .

1. Proof. To show is continuous at 0, we must show that . We show this limit is zero directly from the epsilon-delta definition of the limit, i.e., given arbitrary positive , we have whenever . Let be an arbitrary number greater than 0, . Then, let . Using the given inequality we have,

Thus, is continuous at

2. Proof. First, using the given trig identity we have

Thus,

Thus, cosine is continuous at 0.

3. Proof. Finally, to show sine and cosine are continuous for all , we show that , and . First, we recall the formulas for the sine and cosine of a sum,

So, we compute the limits

Therefore, sine and cosine are continuous for all