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In the text "Elementary Real Analysis The Theory of Calculus", I'm having trouble verifying my attempted proof that the function in $(1)$ is continuous via a $\delta$-$\epsilon$ approach.

Could anyone help to verify my proof, and/or point out errors I made, with suggestions to correct them?

$(1)\;\;$ Statement I need to prove: $\,f(x)$ is continuous for all $x$ in $\mathbb{R}$, where

$$f(x) = \sin^2(x) + \cos^6(x)$$

$\text{Lemma}:$ In order to unpack the notion that $f(x)$ is continuous for $x$ within $\mathbb{R}$, one needs the following theorem in $(2)$.

$(2)\;\;$
Let $f$ be a real-valued function whose domain is a subset $\mathbb{R}$. Then $f$ is continuous at $x_{o} \in \text{dom}(f)$ if and only if for each $\epsilon > 0$ there exists some $\delta > 0$ such that $|x-x_{o}| < \delta$ implies $|f(x)-f(x_{o})| < \epsilon$ for every $x \in \text{dom}(f)$.

$(3)\;\;$ Utilizing the recent results developed in $(2)$, one can make the following observations:

$$|f(x)-f(x_{o})| = |1-\cos^2(x) + \cos^2(x) - (1 - cos^2(x) + cos^{6}(x_{o})|$$ $$|1-cos^{2}(x) - 1-cos^2{(x_{o}})|$$ $$|1| \cdot |x - \cos^{2}({x_{o}})|$$

So it's clear that: $$|x - \cos^{2}({x_{o}})| < \frac{\epsilon}{|1|} \text{implies} \, |1| \cdot |x - \cos^{2}({x_{o}})| < |1| \cdot \frac{\epsilon}{|1|} = \epsilon$$

Zophikel
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  • This is a fine exposition. Only suggest is to say, explicitly, in a sentence at the bottom of your post, something along the lines of "Can someone help me to verify my work, and/or, offer feedback on what I might be missing?" So that you have a real question (which ends in a question mark.) – amWhy Jul 04 '17 at 15:25
  • Thanks @amWhy :-), i'll have to reedit this post in a couple of minutes. – Zophikel Jul 04 '17 at 15:29
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    Please review my edits. I formed the explicit question I suggested, and did minor tweaks (a line break/ added spacing, deleting a couple of awkward comments, but it is still your question. If you have other changes you want to make, feel free, and if you want to undo my edits, to start over, all you need to do is click on the link "edited x min ago" (It appears right over the identicon of the most recent editor, which you'll see in the bottom central part of the post. – amWhy Jul 04 '17 at 15:43
  • Yeah i'm looking at the edits now, I would have edited asap normally but there's other stuff i'm working on, i'll have to learn how to do formal latex. – Zophikel Jul 04 '17 at 15:44
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    You did fine with the formatting. Feel free to edit/tweak, add additional work or thoughts, etc. – amWhy Jul 04 '17 at 15:46
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    The statement of the Lemma is badly mangled. It should be: Let $f$ be a real-valued function whose domain is a subset of $\mathbb{R}$. Then $f$ is continuous at $x_{0} \in \text{dom}(f)$ if and only if for each $\epsilon > 0$ there exists some $\delta > 0$ such that $\lvert x - x_{0}\rvert < \delta$ implies $\lvert f(x) - f(x_{0})\rvert < \epsilon$ for every $x \in \text{dom}(f)$. – murray Jul 04 '17 at 15:49
  • Ahhh ok yeah the Theorem within the $\text{Lemma}$, is badly mangled i'll have to reedit :). – Zophikel Jul 04 '17 at 15:52
  • The first line of the display in (3) is also mangled, as there are unbalanced parentheses. Are you trying to put there, on the right-hand side of the equality, an absolute value of differences, or a difference of absolute values, or an absolute value of difference of absolute values? – murray Jul 04 '17 at 16:02
  • In $(3)$, I was attempting to express the absolute value of differences, looking back at the question the proof may have gotten mangled due to typo's sorry :(. – Zophikel Jul 04 '17 at 16:06
  • You can probably use the addition laws for sine and cosine somewhere in the proof. – mathreadler Jul 04 '17 at 16:08
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    First use the triangle inequality to write $|f(x) - f(x_0)| \leq |\sin^2(x) - \sin^2(x_0)| + |\cos^6(x) - \cos^6(x_0)|$. Then you can factor $|\sin^2(x) - \sin^2(x_0)| = |\sin(x) - \sin(x_0)|\cdot |\sin(x) + \sin(x_0)| \leq 2 |\sin(x) - \sin(x_0)|$ since $|\sin(x)| \leq 1$. You can do a similar thing for the $\cos$ term. These manipulations reduces the problem to the continuity of $\sin(x)$ and $\cos(x)$. $\epsilon-\delta$ derivations for this exists on this site, try searching if you don't know these. – Winther Jul 04 '17 at 16:13
  • Ahh ok looking back at my proposed solution I think I made some mistakes by not employing the triangle inequality. – Zophikel Jul 04 '17 at 16:21
  • I don't understand what you have done in your approach. Where do you use that $|x-x_0| < \delta$? Your last equation says that $|x - \cos^2(x_0)| < \epsilon$. I don't see why and how you get this and even why this statement should prove anything given what you have written above. For a more minor critique: (2) is not really a theorem, it's the definition of continuity (when slightly rephrased as done in one of the comments above). – Winther Jul 04 '17 at 16:38
  • In the text I was reading what was in $(2)$, was stated as a theorem. – Zophikel Jul 04 '17 at 16:41
  • I see. Well there are many other definitions of continuity and if the book uses another one then this could be a theorem so just forget I mentioned it. – Winther Jul 04 '17 at 16:53
  • All right now for the steps I performed for my purposed solution, I looked at a general proof:http://mathworld.wolfram.com/Epsilon-DeltaProof.html. And attempted to adapted what was used. – Zophikel Jul 04 '17 at 16:55
  • As a correction that will have to be noted later I should have added this important remark: For every $\epsilon > 0$ there is a $\delta > 0$ such that whenever $|x-x_{o}| < \delta$, then $|f(x)-f(x_{o})| < \epsilon$. to clarify how $|x-x_{o}| < \delta$ was used, other then this I'm not sure on where to go correcting the mistakes within this proof. – Zophikel Jul 04 '17 at 17:01
  • OK, that page only works for $f(x) = ax+b$. I would suggest you start with a simpler example before jumping into this problem (which requires a bit more) just to learn how it works. For example you can try proving that $f(x) = x^2$ is continuous: https://math.stackexchange.com/questions/264081/prove-x-mapsto-x2-is-continuous – Winther Jul 04 '17 at 17:16
  • All right I will I've gotten the basic Intution on how delta expulsion arguments work – Zophikel Jul 04 '17 at 17:22

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