Can anybody check this proof for a Sine limit?

[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).

Homework Statement

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

Homework Equations

Given below:

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,

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

 

[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.

 

[pasmith]

Homework Statement

Prove [itex]\lim_{x \to 0} \frac{x}{\sin^2(x) + 1} = 0 [\itex]<br /> <br /> <h2>Homework Equations</h2><br /> Given below:<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Let<br /> $|x| < 1 \implies -1 < x < 1$<br /> <br /> $\sin^2(-1) + 1 < \sin^2(x) + 1 <\sin^2(1) + 2$[/itex]

[itex] <br /> 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. <br /> <br /> (You need to remember that $\sin(-x) = -\sin(x)$, and thus $\sin^2(-x) = \sin^2(x)$ for every $x \in \mathbb{R}$.)<br /> <br /> Also your lower bound is false: for sufficiently small $|x|$, $1 + \sin^2(x) < 1 + \sin^2(-1)$ because $\lim_{x \to 0} 1 + \sin^2(x) = 1 < 1 + \sin^2(1)$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} > \frac{1}{\sin^2(1) + 1}$ </div> </div> </blockquote><br /> There's an error here; your upper and lower bounds on $\frac{1}{\sin^2(x) + 1}$ are actually equal.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} \\ \implies \frac{1}{|\sin^2(-1) + 1|} > \frac{1}{|\sin^2(x) + 1|} \\ \implies \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$<br /> <br /> $(1) |x| < \delta_1$<br /> <br /> $(2) \displaystyle \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$<br /> <br /> $(3) \displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1} {|\sin^2(-1) + 1|}$<br /> <br /> Finally,<br /> <br /> $\epsilon(\sin^2(-1) + 1) = \delta_1$<br /> <br /> Therefore,<br /> <br /> $\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$ </div> </div> </blockquote><br /> Your argument is hard to follow. You need to use more connecting phrases and fewer implication symbols.<br /> <br /> You can make your task much, much easier by observing that $1 + \sin^2(x) \geq 1$ so that $$\left| \frac{x}{1 + \sin^2 x}\right| = \frac{|x|}{1 + \sin^2(x)} \leq |x|.$$[/itex]

 

[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?

 

[vela]

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,

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

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.

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

Which would complete the proof?

 

[PeroK]

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?

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.

 

[Amad27]

Sorry, What is the difference between implies and then?

 

[pasmith]

Sorry, What is the difference between implies and then?

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 "$(P \implies Q) \implies R$".