Squeeze theorem?

Question

can someone please explain the squeeze theorem to me. this theorem is sometimes known as the pinch theorem and sandwich theorem. please try to explain it in simplest terms.

Thanks.

Answer 1

Generally this is a theorem about limits. Suppose you have a sequence, x(n). Also suppose x(n) is bounded between two other sequences, a(n) and b(n). That is a(n) is less than or equal to x(n) is less than or equal to b(n) for all n. If the limits of a(n) and b(n) as n-> infinity are equal to say L (lim{a(n)} = lim{b(n)} = L), then lim{x(n)} = L. The picture is that x(n) is being "squeezed" or "sandwiched" between a(n) and b(n).

This can also be stated in terms of functions. If f, g, h are a funtions such that g is less than f is less than h near a point y, and the limits of g(t) and h(t) as t-> y are both equal to L, then the limit of f(t) as t -> y is also equal to L.

Answer 2

A good example of the squeeze theorem are the functions x^2, -x^2, and x^2*sin(1/x)

You can show (or see by looking at the graph) that x^2 is always greater than -x^2.

Now, as for x^2*sin(1/x) (If you haven't had trig yet, forgive me for using this function. The sandwich theorem is usually given in calculus classes, so I'm assuming that you're familiar with trig)

Anyway, x^2*sin(1/x) will always be less than x^2 and always be greater than -x^2. This is because the MOST that sinx can be is 1, which makes it x^2*(1), and the LEAST that sinx can be is -1, which makes it x^2*(-1) = -x^2. So, x^2*sin(-1/x) will always be between x^2 and -x^2.

The sandwich theorem comes into play if you're trying to find the limit as x --> 0 of x^2*sin(1/x). You can't just plug in zero, because you'll get something that's undefined...sin (1/0). So, since the lim as x --> 0 of x^2 is 0, and the lim as x -->0 of -x^2 is 0, and x^2*sin(1/x) is always in between x^2 and -x^2, then the lim as x -->0 of x^2*sin (1/x) has to be 0 by the Sandwich Theorem.

Answer 3

Squeeze theorem
From Wikipedia, the free encyclopedia
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In calculus, the squeeze theorem (also known as the pinching theorem or the sandwich theorem) is a theorem regarding the limit of a function. The theorem asserts that if two functions approach the same limit at a point, and if a third function is "squeezed" ("pinched", "sandwiched") between those functions, then the third function also approaches that limit at that point.

The squeeze theorem is a technical result which is very important in proofs in calculus and mathematical analysis. It is typically used to confirm the limit of a function via comparison with two other functions whose limits are known or easily computed. It was first used geometrically by the mathematicians Archimedes and Eudoxus in an effort to compute π, and was formulated in modern terms by Gauss.

In Italian, the squeeze theorem is also known as the two carabinieri theorem. The idea is that two carabinieri are holding a prisoner in the middle.

The sandwich/squeeze theorem has no relation to the ham sandwich theorem.

example while proving the important limit sin theta/theta =1 theta tending to zero
lt theta tending tozero 1>lt theta ->0sin theta/theta>lt theta ->0 cos theta
1>lt.theta ->0 sin theta /theta>1
since the limit sintheta /theta theta ->0 is squeezed bet. 1 and 1 we conclude it is = 1

Answer 4

Let's say you have 3 functions (A, B, C). If you know that for every real value of x, A >= B >= C. In otherwords, you have two boundaries on B (A is above it, C is below it). If you get to some value of x for which A = C, even if you don't know what it is for B, you've already said that B must be >= C and <= A. The only value that both of those hold for is that same value. Hope this helps.

Answer 5

i dont know if this will help you but ill try to explain it...

given 3 functions a, b, and c; where c is in between a and b (given that a>b) and the limit as x goes to inifiny of both a and b are the same (ex. L-which represents some number) then the limit of c is also L.

i hope i just didnt confuse you more.
goodluck!

Answer 6

-x ≤ xsin(1/x) ≤ x, since -1 ≤ sin(1/x) ≤ 1 Taking limit x->0 gives the answer: xsin(1/x) = 0 ------ To the answerer above me, You missed the '=' sign.