Can anybody check this proof for a Sine limit?

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1. Nov 5, 2014

Amad27

Mod note: Fixed the LaTeX. The closing itex tag should be /itex, not \itex (in brackets).
I find it easier to use # # in place of itex, or  in place of tex (without the extra space).

1. The problem statement, all variables and given/known data

Prove $\lim_{x \to 0} \frac{x}{\sin^2(x) + 1} = 0$

2. Relevant equations
Given below:

3. The attempt at a solution

Let
$|x| < 1 \implies -1 < x < 1$

$\sin^2(-1) + 1 < \sin^2(x) + 1 <\sin^2(1) + 2$

$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} > \frac{1}{\sin^2(1) + 1}$

$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} \implies \frac{1}{|\sin^2(-1) + 1|} > \frac{1}{|\sin^2(x) + 1|} \implies \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$

$(1) |x| < \delta_1$

$(2) \displaystyle \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$

$(3) \displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1} {|\sin^2(-1) + 1|}$

Finally,

$\epsilon(\sin^2(-1) + 1) = \delta_1$

Therefore,

$\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$

Last edited by a moderator: Nov 5, 2014
2. Nov 5, 2014

gopher_p

This looks more like the scratch work you do in order to figure out the right $\delta$ given an arbitrary $\epsilon$. The proof is the part where you demonstrate that the $\delta$ that you've found works; i.e. the part where you say something along the lines of, "let $\delta=...$" usually occurs somewhere in the neighborhood of the second sentence, not at the very end. Most proof readers care less (or not at all) about how you found $\delta$ than they do about whether or not it actually works.

Also note that, no matter what $x$ is (as long as $x$ is a real number), $\sin^2 x+1\geq1$. So $\frac{1}{\sin^2 x+1}\leq1$.

P.S. The first sentence of nearly any $\epsilon-\delta$ proof is, "Fix $\epsilon>0$," or something along those lines.

Last edited: Nov 5, 2014
3. Nov 5, 2014

pasmith

It feels more natural to let $|x| < \frac \pi 2$, which is the nearest maximum of $\sin^2(x)$. Then you won't have $\sin^2(1) = \sin^2(-1)$ cluttering up your argument.

(You need to remember that $\sin(-x) = -\sin(x)$, and thus $\sin^2(-x) = \sin^2(x)$ for every $x \in \mathbb{R}$.)

Also your lower bound is false: for sufficiently small $|x|$, $1 + \sin^2(x) < 1 + \sin^2(-1)$ because $\lim_{x \to 0} 1 + \sin^2(x) = 1 < 1 + \sin^2(1)$.

There's an error here; your upper and lower bounds on $\frac{1}{\sin^2(x) + 1}$ are actually equal.

Your argument is hard to follow. You need to use more connecting phrases and fewer implication symbols.

You can make your task much, much easier by observing that $1 + \sin^2(x) \geq 1$ so that $$\left| \frac{x}{1 + \sin^2 x}\right| = \frac{|x|}{1 + \sin^2(x)} \leq |x|.$$

4. Nov 5, 2014

Amad27

Hello, so if
If we have to prove $|\frac{x}{1+\sin^2(x)}| \le |x|$

In x of real numbers, $1 < 1 + \sin^2(x) < 2$

So $1 \ge \frac{1}{1+\sin^2(x)} \ge \frac{1}{2}$

I see then,

$|x| \ge \frac{1}{1+\sin^2(x)}$

So if $\delta = \epsilon$ and $|x| < \delta \implies |x| < \epsilon$

Which would complete the proof?

5. Nov 5, 2014

vela

Staff Emeritus
Perhaps this is just a typo, but what you wrote certainly isn't true for small $x$. Suppose $x \cong 0$. Then $1+\sin^2 x \cong 1$, so the righthand side is approximately equal to 1, which is bigger than $\lvert x \rvert$ since, by assumption, $x$ is close to 0.

6. Nov 5, 2014

PeroK

Hopefully you've learned from this that it's not a good idea to dive in with ε-δ but is much better to simplify things first.

Why not try to write out a proper, formal proof now? Starting with "Let ε > 0 ...".

Note that your last statement is, in fact, a mixture of "if ... then" and "⇒". You have a choice. You can say:

If $|x| < δ$ then ...

Or

$|x| < δ \ \Rightarrow \dots$

But, it's not quite right to mix these up.

7. Nov 6, 2014

Amad27

Sorry, What is the difference between implies and then?

8. Nov 6, 2014

pasmith

Grammar.

"$P \implies Q$" ("P implies Q") is a complete mathematical statement, which is equivalent to "If P then Q".

"If $P\implies Q$" is an incomplete statement, which must be completed by "then R", with the meaning of "if P implies Q, then R", or in symbolic form "$(P \implies Q) \implies R$".

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