Question

sandwich theorem??
Can you explain it clearly.I dont ge it at all.Thank you so much

Answer 1

you mean about limits?

Answer 2

yes

Answer 3

I say you should have a look at the first example on wiki page. It can't be stated more clearly than that

http://en.wikipedia.org/wiki/Squeeze_theorem

Answer 4

nice picture there too illustrating the idea

Answer 5

Math noton on that pagis confusing me

Answer 6

Suppose you are a plane flying into a SANDWICH. Well, there is bread below you, and bread above you. And it's coming together. Well, there's only ONE place you can end up and that's among the bologna.

So if the BOTTOM bread is at height 800 meters and the TOP bread is at height 800.004 meters, then both YOU and the BOLOGNA (sounds like bolony) must be at about 800.002 meters.

Answer 7

so if f(x) is the bottom bread and g(x) is the top bread, and p(x) is the plane. Then because you are between the breads, we know f(x) < p(x) < g(x).

Now as x goes to infinity, f(x) approaches 800 and g(x) also approaches 800. Then we know that as x approaches infinity, 800 <= p(x) <= 800.

Well there is only one number between 800 and 800, and that's 800! So p(x) = 800.

Answer 8

merit badge please

Answer 9

thank you thank you!

Answer 10

LOL @mareoraft , you are asking for it for quite some time. Thought I would give you one :)

Answer 11

can you solve an example function using this propery

Answer 12

it is most appreciated

Answer 13

Sure, give us the example problem.

Answer 14

lim x approaches 0 [x^2sin(1/x)]

Answer 15

See the first example at wikipedia. It is the same problem!

Answer 16

I read it and didnt understand it there

Answer 17

Tell me what exactly did you not understand ther ?

Answer 18

*there

Answer 19

You just need to know that

-1 < sin x < 1

Answer 20

Multiply x^2 to all the 3

-x^2 < x^2 sin x < x^2

Answer 21

Apply limit x tends to 0

you get

0< (limit x--> 0) x^2 sin x <0

Therefore
(limit x--> 0) x^2 sin x = 0

Answer 22

why is - <sin 1/x <1

Answer 23

Read about sin function

Answer 24

LIM as x approaches 0 OF [x^2sin(1/x)]

-1 <= sin(1/x) <= 1
-x^2 <= x^2*sin(1/x) <= x^2 but now we let x go to 0, and we get...
0 <= x^2*sin(1/x) <= 0

so the answer is 0.

Answer 25

Here http://en.wikipedia.org/wiki/Sine

Answer 26

I know -1<sinx<1 but why -1<sin 1/x<1

Answer 27

it's -1 <= sin(y) <= 1

NOT -1 < sin(y) < 1

Answer 28

I knoq just didnt know how to put the notation here

Answer 29

it doesn't matter what you put in the sin(). nomatter what you put in, its still bounded by -1 and 1.

-1 <= sin(15u^7) <= 1