Does it matter if L + epsilon is not in the range of the function?

[rakeru]

Homework Statement

Use a graphing calculator to find \delta
when

0<|x - \pi/2|<\delta and |sin(x) - 1|<0.2

Homework Equations

I don't think there are any other than the format of the previous information:

0<|x-a|<\delta and |f(x)-L|<\epsilon

The Attempt at a Solution

Okay, so I used the graphing calculator to graph sinx, y=1.2 and y=0.8
The thing is, L + ε turns out to be greater than 1, so I have something like this (i attached a picture).

For \delta I got 0.61. I'm not sure if it's possible, because the line y=1.2 doesn't touch the graph of sinx. Is it possible?

Thank you!

 

[Dick]

Homework Statement

Use a graphing calculator to find \delta
when

0<|x - \pi/2|<\delta and |sin(x) - 1|<0.2

Homework Equations

I don't think there are any other than the format of the previous information:

0<|x-a|<\delta and |f(x)-L|<\epsilon

The Attempt at a Solution

Okay, so I used the graphing calculator to graph sinx, y=1.2 and y=0.8
The thing is, L + ε turns out to be greater than 1, so I have something like this (i attached a picture).

For \delta I got 0.61. I'm not sure if it's possible, because the line y=1.2 doesn't touch the graph of sinx. Is it possible?

Thank you!

That's just fine. Sure, sin(x) never exceeds 1.2, so you don't have to worry about that case.

 

[SammyS]

Homework Statement

Use a graphing calculator to find \delta
when

0<|x - \pi/2|<\delta and |sin(x) - 1|<0.2

Homework Equations

I don't think there are any other than the format of the previous information:

0<|x-a|<\delta and |f(x)-L|<\epsilon

The Attempt at a Solution

Okay, so I used the graphing calculator to graph sinx, y=1.2 and y=0.8
The thing is, L + ε turns out to be greater than 1, so I have something like this (i attached a picture).

For \delta I got 0.61. I'm not sure if it's possible, because the line y=1.2 doesn't touch the graph of sinx. Is it possible?

Thank you!

No.

Edited:
You just need to have $\displaystyle \ L-\varepsilon < f(x) < L+\varepsilon \$ whenever $\displaystyle 0< \left| x-a \right| <\delta$

 

[rakeru]

Uhm, okay. So... who is right?
I don't quite understand what L-ε < f(x) < L -ε means. Do you mean L - ε < f(x) < L + ε ?

 

[Dick]

Uhm, okay. So... who is right?
I don't quite understand what L-ε < f(x) < L -ε means. Do you mean L - ε < f(x) < L + ε ?

I'm not sure what SammyS is on about. If |x-pi/2|<0.61 then |sin(x)-1|<0.2. At least approximately, as your graph is showing you. I don't see any problems with what you did.

 

[SammyS]

I'm not sure what SammyS is on about. If |x-pi/2|<0.61 then |sin(x)-1|<0.2. At least approximately, as your graph is showing you. I don't see any problems with what you did.

Mainly a typo.

I'll fix it.

 

[Dick]

Mainly a typo.

I'll fix it.

Ah, ok so your "No" was answering the "Does it matter" question in the post question and my "Yes" was answering the "Is it possible" question posed by the OP in the post. No conflict.