# Implications of Squeeze Theorem

1. Sep 5, 2009

### honestrosewater

I was just introduced to the http://en.wikipedia.org/wiki/Squeeze_theorem" [Broken] in the second week of Calc 1. This theorem implies that $\lim_{x \to c} f(x)$ exists under certain circumstances, and I want to find out and prove what (at least some of) those circumstances are and what the value of the limit is.

Roughly, my idea so far is to use Squeeze and supremums and infimums to find the one-sided limits, giving me the two-sided limit when they agree. I would like to know if I am messing up somewhere or if anyone has some time-saving hints for me. (This is not homework, so I don't have much time to spend on it.) All functions are real-valued.

Theorem X, rough draft. I need to cover 4 cases: increasing from the left, decreasing from the left, increasing from the right, and decreasing from the right.
(a) If f is bounded and increasing (or decreasing) on [b,c), then $\lim_{x \uparrow c} f(x)$ = the supremum (or infimum) of the image of f on [b,c).
(b) If f is bounded and increasing (or decreasing) on (c,d], then $\lim_{x \downarrow c} f(x)$ = the infimum (or supremum) of the image of f on (c,d].

PROOF

(a-increasing) Let s = the supremum of the image of f on [b,c). The pair of squeezing functions will be g(x) = s and h(x) = ((s - f(b))/(c - b))x, i.e., a horizontal line through s and a secant line to f passing through f(b) and s. That these two functions will squeeze f looks obvious since f is increasing on the interval. And the supremum should exist since I am dealing with subsets of R.

I also need to prove that the limits of my squeezing functions g(x) and h(x) always exist and are equal approaching any c. Both limits exist since g(x) is constant and h(x) is linear and defined everywhere since c > b. Should I note that c > b or is this assumed for an interval? I'm not sure how to prove that the limits are equal.

The other three cases would work similarly.

My last step is applying Squeeze. Is there a one-sided version of Squeeze? if not, I would have to combine the g(x) and h(x) that I get from the left with the ones that I get from the right. If the one-sided limits are equal, g should be the same horizontal line through s, and h should be either a line through s or an absolute value function with vertex s. So combining them should be straightforward.

Will this work? Will f need to be continuous on the interval?

Last edited by a moderator: May 4, 2017
2. Sep 7, 2009

### n!kofeyn

I'm not really for sure what you're talking about. The squeeze theorem says:
If $g(x)\leq f(x)\leq h(x)$ for all x in an interval containing a and
$$\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$$, then $$\lim_{x\to a} f(x) = L$$.

The squeeze theorem gives you exactly which circumstances must be satisfied for the limit of f to exist, and it gives you the value of the limit of f if these circumstances are satisfied! So I'm not for sure why you are trying to find these circumstances or what the value of the limit of f is. Are you trying to prove this theorem? When you are proving something, you don't need to prove what is given (i.e. you don't have to prove that the limits of g and h exist and are equal, you assume that). You also don't need to know anything about the continuity of g, f, or h to apply the squeeze theorem.

3. Sep 7, 2009

### slider142

If you are attempting to prove the Squeeze theorem, one problem is that you assume that f is either increasing or decreasing on some interval, which is not necessarily true and is not implied by any assumption in the theorem.
As an illustration, consider the function:
$$f(x) = \left\lbrace\begin{array}{ll}x^2\sin\left(\frac{1}{x}\right), & x\neq 0\\ 0, & x=0\end{array}$$
With respect to the limit as x approaches 0, the squeeze theorem is applicable ($g(x) = -x^2, h(x) = x^2$) and gives the correct limit, and yet in no neighborhood of 0 is f increasing or decreasing, which are the only behaviors considered in your attempt.
The simplest proof is a direct algebraic application of the fact that the limits for g and h exist and are equivalent to the inequality, with no need to go lower than limit theorems (no need for delta-epsilon or infimum/supremum considerations).

Last edited: Sep 7, 2009
4. Sep 7, 2009

### honestrosewater

I am not trying to prove Squeeze. I am trying to find a pair of functions that will always squeeze certain f. But I messed up several things. First, it's not enough that f be increasing or decreasing. I am concerned with the concavity of f, so I care whether the derivative of f is increasing or decreasing. I don't know why it was so hard to express what my brain was doing here.

Second, Squeeze requires an open interval that contains the point whose limit you're trying to find. That error was just sloppiness.

Third, an absolute value function, which is what I would have ended up with in some cases, is continuous but not differentiable at its vertex. This error was ignorance.

So my idea needs an overhaul, but this has still been a good learning experience.

/

5. Sep 7, 2009

### slider142

Hmm, are you assuming the limit exists? If so, what about "given the limit as x approaches a of f(x) to be L, consider g(x) = L + |x - a|f(x) and h(x) = L - |x - a|f(x)." The inequality follows from the absolute value, and they squeeze the function pretty well. Did you need h and g to be differentiable as well?

6. Sep 7, 2009

### Hurkyl

Staff Emeritus
For the record, the following are true statements:

If f is bounded and increasing on [b,c), where b<c, then $\lim_{x \uparrow c} f(x)$ is indeed the supremum of the image under f of [b,c).​
(note: you can replace [b,c) with (b,c))

If $f \leq g \leq h$ on the interval (b,c), where b<c, and $\lim_{x \uparrow c} f(x) = \lim_{x \uparrow c} h(x) = L$, then $\lim_{x \uparrow c} g(x) = L$​
(i.e. one-sided squeeze is a theorem)

7. Sep 7, 2009

### n!kofeyn

I guess I'm still confused. By certain f do you mean given any f you want to explicitly find functions g and h that squeeze its limit at a point? What about sin(1/x)? You can in no way squeeze it to a limit at 0.

8. Sep 7, 2009

### honestrosewater

No, by "certain", I do not mean "any". I mean f that meet certain requirements, such as being bounded and nonoscillating approaching c.

Sorry, I thought it was a straightforward desire. Squeeze says that one way to find limits of f is to find appropriate pairs of squeezing functions. Right? I want to know if, in some cases, there is a general way to define this pair of squeezing functions, i.e., I want to define g and h in terms of f and the interval.

But if I can define g and h this way, I don't even need them, only their common limit. Right? In my example above, their common limit would have been s, so I can skip thinking about g and h altogether and just say that the limit of f approaching c is s under these circumstances.

Maybe this seems like a roundabout way of doing things. It was just meant as a learning exercise.

Well, if the limit of f approaching a doesn't exist, my pair of squeezing functions won't exist either. So, stupid question: what is the relationship between the limit of f(x) as x approaches c and whether f is differentiable at c?

9. Sep 8, 2009

### Elucidus

The squeeze theorem is applicable in very broad situations regardless on the nicety of the function f.

Note, for

$$f(x) =\left\{ \begin{array}{rl} x\sin(1/x), & \text{if }x \neq 0 \\ 0, & \text{if }x = 0 \end{array}$$

$$\lim_{x \rightarrow 0} f(x) = 0$$ by the Squeeze Theorem.

Similarly if w(x) is Weierstrass' pathological function which is continuous everywhere, but not differentiable anywhere then

$$\lim_{x \rightarrow 0} x \cdot w(x) = 0$$ also by the Squeeze Theorem.

In general there is no great way to fabricate the inferior and superior bounding functions for any given function f. It'd be nice to have such a way, but alas, things are not to be.

If f is sufficiently nice to be twice differentiable, or monotonic, or whatnot, then there are probably other ways to find the limit (i.e. continuity).

--Elucidus