Test II like exercises

Exercise 4.4.1 Let $f(x) = x^{3}$. Show $f$ is continuous on $\mathbb{R}$.

Proof. Let $x, x_{0} \in \mathbb{R}$, $\epsilon > 0$ and
$$\delta = \textrm{ min} \{1, \frac{\epsilon}{3(x_{0} + 1)^2} \},$$ satisfy $| x - x_{0}| < \delta$. Then $|f(x) - f(x_{0})| = |x^3 - x_{0}^{3}| = |x - x_{0}|\cdot|x^2 + xx_{0} + x_{0}^2|$ Note that $x \in (x_{0} - 1, x_{0} + 1)$, by our choice of $\delta$. It follows that $|x^2 + xx_{0} + x_{0}^2| < (x_{0} + 1)^2 + (x_{0} + 1)^2 + x_{0}^2 < 3(x_{0} + 1)^2$. Substituting in we find that $|x - x_{0}|\cdot|x^2 + xx_{0} + x_{0}^2| < \delta\cdot3 (x_{0} + 1)^2 \leq \frac{\epsilon}{3 (x_{0} + 1)^2} \cdot3 (x_{0} + 1)^2 = \epsilon$. So $|f(x) - f(x_{0})| < \epsilon$. $\square$.

#5 part a on the Review Sheet
Prove that $f(x) = 3x + 1$ is uniformly continuous on $\mathbb{R}$.

Proof.
Let $\epsilon > 0$ and choose $\delta = \frac{\epsilon}{3}.$
Then, $ 0 < |x-c| < \delta$ implies,
$$|f(x) - f(c)| $$
$$= |(3x+1)-(3c+1)| $$
$$= 3|x-c| < 3\delta = 3(\frac{\epsilon}{3}) = \epsilon$$

Thus, since $|f(x) - f(c)|$ we know $f(x)$ is uniformly continuous on $\mathbb{R}$.

#5 b on review sheet

We will show that $lim_{x \rightarrow c} ((f(x)-f(c))=0$ or equivalently $lim_{x \rightarrow c} f(x)=f(c).$

So, $lim_{x \rightarrow c}(f(x)-f(c))=lim_{x \rightarrow c} (\frac{f(x)-f(c)}{x-c})(x-c)=lim_{x \rightarrow c}\frac{f(x)-f(c)}{(x-c)} \cdot lim_{x-c}(x-c).$

Since $lim_{x \rightarrow c}(x-c)=0$, then $f'(x) \cdot 0=0.$

Look good?

Does anyone have any insight into Exercise 4.5.2 in the book?

For part b I believe it is true since a continuous function that goes through an open interval contains all of the points in the interval and thus can be written as a the open set of those points.

I would like to know what others think on my reasoning and also parts a and b!

b is the part that I'm not really too sure about. a is false, because $f(x)=\frac{1}{x}$ goes from bounded on $(0,1)$ to the unbounded interval of $(1,\infty)$. Part c is true because of theorem 4.4.2 (Preservation of Compact Sets) to eventually show that the compact set is an interval. b i think would be false because of the counter example $f(x)=x^{2}$ goes from $(-1,1)$ to $[0,1)$. I'm not positive, though, so I'm definitely open for thoughts and feedback.

Ah! I think I understand the questions more clearly after seeing your reasoning...I think I will have to rethink my response to b.

#4 part b. on the review guide

A continuous function $f: [0,1] \mapsto \mathbb{R} $and a cauchy sequence $x_n$ such that $f(x_n)$ is not cauchy.
I am struggling to understand how this one is impossible can anyone explain this?

@lhouse1, in your proof for #5 part a, I am not sure we need the $0< \left|x-c\right|$ for uniform continuity since we don't necessarily need the condition that $x\neq c$. I may be wrong and it may not even matter.....

Here is what I have found out about #4 part b from the in class problems:

If you have a Cauchy sequence $x_n$ in $[0,1]$, then it has a limit in $[0,1]$ since this interval is closed.

If you let the $\lim x_n = x$, then by continuity $f(x) = \lim f(x_n)$. Since $f(x_n)$ always converges, it will always be Cauchy.

I know 4c is possible on the review sheet and I tried to think of an example and this is what I came up with:

For it to have a maximum but not a minimum, I figured it had to be something parabolic. Initially I thought a $\sin(x)$ or $\cos(x)$ function but I'm lazy and thought that might be too complicated. Then I figured why not work with just a parabola? So I'm thinking maybe something like $f(x) = x-x^2$. That way the roots are $0$ and $1$ and it's maximum value is at its vertex $\left(\frac{1}{2}, \frac{1}{4} \right)$.

Not sure about reasoning(s) or examples for the other ones though!

Simple, Elegant, Perfect

@qkhan I think you had a slight typo. I'm pretty sure you are discussing 4c and if so I agree.

You are correct, I did mean 4c. 4b isn't possible. My bad, thanks for pointing it out!

Good job, just one minor thing, we only care that $0 < |x - c|$ when we are dealing with functional limits. for continuity it's perfectly fine to have $x = c$

any ideas for the types of problems that may be on the friday exam

I asked Mark and he said #3 on the study guide would be a fair game..

But of course he would😏

I think we should maybe focus on Cantor Set stuff, more open/closed set stuff...I don't know. Any other ideas?

Am I correct i thinking we won't have another round of definitions on this next test?

last exam we only had definitions on the wednesday part, however we had a longer list this time and only a few appearing on the first part so it wouldnt surprise if there were more