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

1. Feb 3, 2014

### rakeru

1. The problem statement, all variables and given/known data
Use a graphing calculator to find $\delta$
when

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

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

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

3. 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!

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2. Feb 3, 2014

### Dick

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

3. Feb 3, 2014

### SammyS

Staff Emeritus
No.

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

Last edited: Feb 3, 2014
4. Feb 3, 2014

### 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 ???

5. Feb 3, 2014

### Dick

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.

6. Feb 3, 2014

### SammyS

Staff Emeritus
Mainly a typo.

I'll fix it.

7. Feb 3, 2014

### Dick

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.