Sandwich Theorem

[aron101782]

Sandwich Theorem and trigonometric functions

I have a question regarding the sandwich Theorem.

If

cost<sint/t<1

such that

LIM(t,0)cost=1 and LIM(t,0)1=1

by sandwich theorem we have

LIM(t,0+)sint/t =1
(In other words as t to 0 from the right)
And since sint/t is an even function
(Email me at aronmatthew@hotmail.com for proof of this)

Then
LIM(t,0)sint/t=1
(from both sides)

Wouldn't it be safe to assume this before showing that this is an even function simply based on the sandwich theorem.

 

[HallsofIvy]

Yes, the "sandwich theorem" says that "\(\displaystyle f(x)\le g(x)\le h(x)\) for all x, that \(\displaystyle \lim_{x\to a}f(x)= L\), \(\displaystyle \lim_{x\to a} h(x)= L\) then \(\displaystyle \lim_{x\to a} g(x)= L\)". That is true whether f, g, and h are "even" or "odd" functions or neither or both. That is, it has nothing to do with being an even function.

(I before had "A" rather than the last "L"! That silly error was pointed out to me and I have corrected it.)

 

[JeffM]

Yes, the "sandwich theorem" says that "\(\displaystyle f(x)\le g(x)\le h(x)\) for all x, that \(\displaystyle \lim_{x\to a}f(x)= L\), \(\displaystyle \lim_{x\to a} h(x)= L\) then \(\displaystyle \lim_{x\to a} g(x)= A\)". That is true whether f, g, and h are "even" or "odd" functions or neither or both. That is, it has nothing to do with being an even function.

If \(\displaystyle A \ne L\), I'd very much like to see a rigorous proof of this theorem.

Of course, if it is necessarily true that \(\displaystyle A = L\),

the theorem could be stated more intuitively.

 

[HallsofIvy]

I have edited my response to replace that "A" with the correct "L". Thanks for catching that, JeffM.

 

[aron101782]

I'm sorry I forgot to put that 0<=t<=pi/2

f(t)=sint=y/r
f(-y)=-y/r

Such that sint is an odd function

Therefore

sint/t is even and the limit is the same from both sides



-Aron