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I'm trying to prove a special trig limit, which is...

$$\lim_{x \rightarrow 0} \frac{1 - \cos{x}}{x}=0$$

So far, this is what I have (and I'll explain where I'm confused) Using the squeeze theorem, $h(x) \leq f(x) \leq g(x)$ $$-x^2 + 1 \leq \cos{x} \leq 1 $$ $$-x^2 + 1 - 1 \leq \cos x - 1 \leq 1 - 1$$ $$-x^2 \leq \cos{x} - 1 \leq 0$$ $$0 \leq 1 - \cos{x} \leq x^2 $$ $$0 \leq \frac{1- \cos{x}}{x} \leq x$$

Since limit of $0$ and $x$ equals zero (as $x$ approaches zero), so does $\displaystyle{\frac{1-\cos{x}}{x}}$.

My first confusion, is when I try to graph the last line as separate functions In quadrant $3$ and $4$, it holds up. However, in quadrant $1$ and $2$, it becomes $\displaystyle{x \leq \frac{1-\cos{x}}{x} \leq 0}$.

I'm not sure if this is allowed in squeeze theorem, but I'm a tad bit confused.

I have another guess as to why the end result is incorrect because $-x^2 + 1 \leq \cos{x} \leq 1$ are not the correct "sandwich" functions. $f(x) = \cos{x}$ touches $g(x) = 1$ in more than one spot. Every diagram of squeeze theorem I've seen, the sandwich functions only touch $f(x)$ at one spot. Is this a criteria I'm unaware of for picking $h(x)$ and $g(x)$?

PS, I know I could have used $h(x) = -x^2 + 1$ and $g(x) = x^2 + 1$, but I'd still like to know what I did wrong up top, please.

Thanks for any helps, guys/gals/automatons

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  • $\begingroup$ PS, I don't know how to use math symbols here.... I'm new. Does someone have a youtube link or something for a quick tutorial? $\endgroup$ – Astro Jun 2 '14 at 0:56
  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – user137794 Jun 2 '14 at 0:56
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    $\begingroup$ @Astro while you're still getting used to the equation editor script, a quick and dirty trick is to click the edit button of other people's posts and look at their's. $\endgroup$ – David H Jun 2 '14 at 1:00
  • $\begingroup$ It might be easier to multiply top and bottom by $1+\cos x$. Alternately, note that $1-\cos x=2\sin^2(x/2)$. For your way, there is no need to worry about touching at more than one spot. It would not make any difference to the argument, and anyway near $0$ there is only one spot. $\endgroup$ – André Nicolas Jun 2 '14 at 1:02
  • $\begingroup$ Just type everything in dollar signs to do math symbols. Also there are some specific signs like \frac{}{}. $\endgroup$ – user7000 Jun 2 '14 at 1:18
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It's not exactly correct to go from

$$0\le1-\cos x\le x^2$$ to $$0\le{1-\cos x\over x}\le x$$ because dividing through by $x$ reverses the inequalities if $x$ is negative. What is OK is to conclude

$$0\le\left|{1-\cos x\over x}\right|\le |x|$$

The squeeze theorem still applies.

However, where did the opening inequality, $-x^2+1\le\cos x$, come from?

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  • $\begingroup$ Oooo that actually makes sense now! It would explain why the h(x) and g(x) changes position in the inequality when it transfers from quadrant 3 and 4 to quadrant 1 and 2. Thanks! The "-x^2 + 1 <= cosx" came from me just selecting a sandwich function. I needed a function below cosx that touches it at x = 0. $\endgroup$ – Astro Jun 2 '14 at 15:50
  • $\begingroup$ @Astro, I'm glad to help. As for the $-x^2+1$, it does indeed lie below $\cos x$, but how do you know that it does? When you're first learning to prove things about limits, it's important to keep careful track of what you are allowed to assume. $\endgroup$ – Barry Cipra Jun 2 '14 at 16:11
  • $\begingroup$ I just kind of made an educated guess and checked it with my graphing calculator. I've actually taken calculus before but I'm revisiting it because I'm interested in taking my GRE and I was examining some harder problems I couldn't get as a kid :/ $\endgroup$ – Astro Jun 3 '14 at 0:03
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I would recommend that you use L'Hospital's Rule. That's probably easier, since you can get rid of the denominator instantly. $$\lim_{x \to 0} \frac{1-\cos x}{x} \rightarrow \lim_{x \to 0} \frac{\sin x}{1} = 0$$

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  • $\begingroup$ Although this works, I think the author is trying to prove it with the squeeze theorem. $\endgroup$ – Rocket Man Jun 2 '14 at 1:48
  • $\begingroup$ I was :) But being reminded that this is also an option is appreciated. thanks! $\endgroup$ – Astro Jun 2 '14 at 15:56
  • $\begingroup$ It certainly is an option, as long as you prove that d(-$\cos x$)/dx = $\sin x$. Any ideas on how to go about doing that? $\endgroup$ – John Joy Jun 2 '14 at 17:19
  • $\begingroup$ Yeah I can't recall the exact steps, but I know it has something to do with the trig identities. I'm so bad at those >_< $\endgroup$ – Astro Jun 3 '14 at 0:04
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The reason your inequality fails to hold for negative $x$ is because you divided by $x$, so when $x$ is negative, the inequality signs will be reversed. You could take your third line $$ 0 \le 1 - \cos x \le x^2$$ and argue that, since all of the terms are nonnegative, $$ 0 \le \left|1 - \cos x\right| \le \left|x^2\right|.$$ Then divide both sides by $|x|$ to obtain $$0 \le \frac{|1 - \cos x|}{|x|} = \left|\frac{1-\cos x}{x}\right| \le |x|$$ so that $$\lim\limits_{x \to 0} \left|\frac{1-\cos x}{x}\right| = 0.$$

This also proves that $\lim\limits_{x \to 0} \frac{1-\cos x}{x} = 0$.

(Intuitively, because the only numbers with magnitude close to zero are also close to zero. You can also prove this in a more general setting using the sandwich theorem: for any function $f(x)$, $-|f(x)| \le f(x) \le |f(x)|$, so if $|f(x)| \to 0$ as $x \to c$, then $f(x) \to 0$ as $x \to c$, as well.)

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  • $\begingroup$ This makes sense. Thanks for the "intuitively" comment below. It's good to know that as well $\endgroup$ – Astro Jun 2 '14 at 15:56
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I think what you are overlooking is that n going from the fourth line of your solution to the fifth, you are dividing through the inequality by $x$... right?

But if $x$ is negative, dividing by $x$ should reverse the direction of the inequality. So you end up with two inequalities:

$$0 \leq \frac{1-\cos x}{x} \leq x $$ for $x \geq 0$, and

$$x \leq \frac{1- \cos x}{x} \leq 0 $$ for $x < 0 $.

Now compute the left- and right-hand limits separately, using the squeeze theorem together with the above inequalities.

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  • $\begingroup$ Thanks! The restrictions makes a lot of sense. I feel silly now :P $\endgroup$ – Astro Jun 2 '14 at 15:57
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Here's my simplification: $$\lim_{x \rightarrow 0} \frac{1 - \cos{x}}{x}= \lim_{x \rightarrow 0} \frac{1 - \cos{x}}{x}\cdot\frac{1 + \cos x}{1 + \cos x} = \lim_{x \rightarrow 0} \frac{\sin^2 x}{x(1 + \cos x)} $$ $$= \frac{\sin 0}{1 + \cos 0}\cdot \lim_{x \rightarrow 0} \frac{\sin x}{x} = 0\cdot\lim_{x \rightarrow 0} \frac{\sin x}{x}=0$$

All that's left is to use the squeeze theorem to prove that $$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ The proof would be similar to this proof Find the limit $\displaystyle \lim_{x \to 0^+} (\sin x)^\frac1{\ln x}$

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