Squeeze Theorem: Prove Limit of x→0 for Inequality

[Saladsamurai]

Homework Statement

It can be shown that the inequalities

$$1-\frac{x^2}{6}<\frac{x\sin x}{2-2\cos x}<1$$

hold for all values of x close to 0. What, if anything, does this tell you about

$$\lim_{x\rightarrow 0}\frac{x\sin x}{2-2\cos x}$$ ?


Well. I know from looking at the answer that the limit is indeed 1. But what I am confused about is how the strict inequality affects things, or how it does not?

I mean, the Squeeze Theorem is written if $g(x)\le f(x)\le h(x)$ ...and so on...

So how does this affect things?

 

[Saladsamurai]

Okay maybe I can answer this. I found a corollary (of sorts) of the Squeeze Theorem that says that if $f(x)\le g(x)$ for all x in some open interval containing c, except maybe at x = c itself, and the limits if f and g bpth exist as x approaches c, then

$$\lim_{x\rightarrow c}f(x)\le\lim_{x\rightarrow c}g(x)$$

I feel like this is related to my question..maybe

 

[mathie.girl]

Homework Statement

It can be shown that the inequalities

$$1-\frac{x^2}{6}<\frac{x\sin x}{2-2\cos x}<1$$

hold for all values of x close to 0. What, if anything, does this tell you about

$$\lim_{x\rightarrow 0}\frac{x\sin x}{2-2\cos x}$$ ?


Well. I know from looking at the answer that the limit is indeed 1. But what I am confused about is how the strict inequality affects things, or how it does not?

I mean, the Squeeze Theorem is written if $g(x)\le f(x)\le h(x)$ ...and so on...

So how does this affect things?

The strict inequalities don't affect anything. The statement of the theorem requires that $g(x)\le f(x)\le h(x)$. Well, if $g(x) < f(x)$, then surely, $g(x)\le f(x)$. The Squeeze Theorem applies even if the functions are equal at some points, but it doesn't require it.

The Squeeze Theorem is very natural when you think about it. It says that if the top and bottom functions both get arbitrarily closer to some number (in this case, 1), then the function that is always in between them must also get arbitrarily close to 1. Otherwise, it would have to break away from the two functions surrounding it. Applying the theorem is even easier. The hard part is determining functions to bound your function (given to you here), proving that they satisfy the inequality $g(x)\le f(x)\le h(x)$ (also given to you, right?) and then determining their limits. But in this case you're given most of the information, and the limits of the bounding functions are easy. So you're left with the very easy step of saying "by the Squeeze Theorem, since $g(x)\le f(x)\le h(x)$, and $$g$$ and $$h$$ go to L, $$f$$ must go to L".

 

[Elucidus]

Okay maybe I can answer this. I found a corollary (of sorts) of the Squeeze Theorem that says that if $f(x)\le g(x)$ for all x in some open interval containing c, except maybe at x = c itself, and the limits if f and g bpth exist as x approaches c, then

$$\lim_{x\rightarrow c}f(x)\le\lim_{x\rightarrow c}g(x)$$

I feel like this is related to my question..maybe

This is not technically a corollary of the Squeeze Theorem, but it is what is classically used to prove the Squeeze Theorem. This fact is tangentially related to your question. The Squeeze Theorem is the most appropriate approach here.

As mentioned earlier $f(x) < g(x) < h(x) \text{ implies } f(x) \leq g(x) \leq h(x)$ so the Squeeze Theorem still holds for the given inequality.

--Elucidus

 

[HallsofIvy]

In particular, if f(x)< g(x) for all x in some neighborhood of a, then
[tex]\lim_{x\rightarrow a} f(x)\le \lim_{x\rightarrow a} g(x)[/itex]<br /> Notice that the "<" becomes "$\le$".[/tex]