How Can I Apply the Squeeze Theorem Without Trig Functions?

[TitoSmooth]

Not sure how to apply the Squeeze Theorem when not given in trig functions.

My question is.

Lim (x^2+1)=1
x→0

not sure what values let's call them K. ie -K≤x^2+≤K.

for instances when I have.

Lim xsin(1/x)=0
x→0

i say. -1≤sin(1/x)≤1

then multiply the whole inequality by x.

-x≤xsin(1/x)≤x

therefore limit as x approaches 0 of sin(1/x)=0


how would I do it for for non trig functions?

 

[tiny-tim]

Hi TitoSmooth!

(try using the X² button just above the Reply box )

You probably wouldn't need it for non-trig (or non-algebraic) functions!

For example, there's no way of applying it to x²+1.

wikipedia has an example, involving two variables:

-|y| ≤ x²y/(x² + y²) ≤ |y|​

 

[vela]

For example, there's no way of applying it to x²+1.

Yes, there is. There are two very simple functions which bound 1+x² from above and from below on the interval [-1,1]. The key is that you only need to concern yourself with an interval containing x=0. You don't have to find a simple function that bounds 1+x² from above for all x.

 

[TitoSmooth]

Yes, there is. There are two very simple functions which bound 1+x² from above and from below on the interval [-1,1]. The key is that you only need to concern yourself with an interval containing x=0. You don't have to find a simple function that bounds 1+x² from above for all x.

Layman terms my man. So I could understand better. Thanks

 

[HallsofIvy]

Layman terms? I can see no technical terms in what Vela wrote, except possibly "interval" and vela defined it: "the interval [-1, 1]".

 

[TitoSmooth]

Layman terms? I can see no technical terms in what Vela wrote, except possibly "interval" and vela defined it: "the interval [-1, 1]".

You don't have to find a simple function that bounds 1+x2 from above for all x.


missread this. I understand now between the closed interval of -1 and 1