# Limits on graphs (epsilon, thelta).

• Sep 19th 2010, 06:03 AM
FullBox
Limits on graphs (epsilon, thelta).
• Sep 20th 2010, 05:16 PM
FullBox
I managed to solve the first problem, which I removed from the first post for neatness/lack of clunkiness. I also made progress on the second problem, though I'm not sure if its in the correct direction or not, and I'm hoping I'll get some help here. I really want to understand this, so when you correct me, please don't just give me an answer but also explain and elaborate on where I went wrong in my work so I can keep it for the record.

a) Here is the conclusion I reached. For all x with 0 < | x - a | < δ, we have | f(x) - f(a) | < ε.
In this case, f(x) = 6, and f(a) = 4. Therefore, would it be right to say that | 6 - 4 | < ε, or | 2 | < ε? That would mean there is no such value for δ, since ε = 2, and | 2 | is not greater than 2.

b) For this second part of the problem, here is what I came up with. Again, the same rule applies:
For all x with 0 < | x - a | < δ, we have | f(x) - f(a) | < ε. f(x) is still 6, and f(a) is still 4. However, the value of ε has changed; now ε = 1. Thus, here is what I did: | 6 - 4 | < ε, or | 2 | < ε. This would also mean that there is no such value for δ since ε = 1, and 1 is NOT greater than 2.

c) For c, I am not sure how to proceed, and would like some sort of nudge in the right direction. A guess here though is that if none of the two above possibilities worked, wouldn't this one too mean that there is no relationship between δ and ε?
• Sep 21st 2010, 03:30 AM
HallsofIvy
Saying "there is NO such value for $\delta$" is saying that the limit does not exist- and it obviously does!

If |f(x)- 6|< 2, then -2< f(x)- 6< 2 so 4< f(x)< 8. Drawing horizontal lines on the graph at y= 4 and y= 8, I see that the line y= 4 crosses the graph at (2, 4) and the line y= 8 crosses the graph at (8, 8). If I were to draw vertical lines at x= 2 and x= 8, I would form a rectangle with the graph completely inside the rectangle. That shows that 4< f(x)< 8, which is the same as |f(x)- 6|< 2, for 2< x< 8. Subtracting 4 from each part, -2< x- 4< 4. Since |-2|< |4|, that will be true as long as |x- 4|< |-2|= 2. We can take $\delta= 2$.

For (b), do the same thing. |f(x)- 6|< 1 is the same as -1< f(x)- 6< 1 and then 5< f(x)< 7. Drawing horizontal lines on the graph at y= 5 and y= 7, I see that the graph crosses y= 5 at (3, 5) and y= 7 at (6, 7). That is, 5< f(x)< 7, which is the same as |f(x)- 6|< 1, as long as 3< x< 6 or -1< x- 4< 2. Since |-1|< 2, that will be true as long as |x- 4|< 1. We can take $\delta= 1$.
• Sep 21st 2010, 03:33 PM
FullBox
Quote:

Originally Posted by HallsofIvy
Saying "there is NO such value for $\delta$" is saying that the limit does not exist- and it obviously does!

If |f(x)- 6|< 2, then -2< f(x)- 6< 2 so 4< f(x)< 8. Drawing horizontal lines on the graph at y= 4 and y= 8, I see that the line y= 4 crosses the graph at (2, 4) and the line y= 8 crosses the graph at (8, 8). If I were to draw vertical lines at x= 2 and x= 8, I would form a rectangle with the graph completely inside the rectangle. That shows that 4< f(x)< 8, which is the same as |f(x)- 6|< 2, for 2< x< 8. Subtracting 4 from each part, -2< x- 4< 4. Since |-2|< |4|, that will be true as long as |x- 4|< |-2|= 2. We can take $\delta= 2$.

For (b), do the same thing. |f(x)- 6|< 1 is the same as -1< f(x)- 6< 1 and then 5< f(x)< 7. Drawing horizontal lines on the graph at y= 5 and y= 7, I see that the graph crosses y= 5 at (3, 5) and y= 7 at (6, 7). That is, 5< f(x)< 7, which is the same as |f(x)- 6|< 1, as long as 3< x< 6 or -1< x- 4< 2. Since |-1|< 2, that will be true as long as |x- 4|< 1. We can take $\delta= 1$.

Thank you very much! I really appreciate your help and you did indeed provide the correct answer. However, there are still several things I do not understand and would really appreciate if you could clarify (I really want to understand and feel comfortable with this topic).

I completely understand how you restricted the function with the vertical and horizontal lines. Epsilon was 2, and so you restricted the function horizontally from y= 6 + ε = 8, and y = 6 - ε = 4. I also understand that the vertical lines restricted the graph from where the function intersected with the aforementioned horizontal lines.

However, in the end, I got a bit confused [it seems I got confused at the easiest part of the explanation]. I know this will come off as a stupid question, but I just want to totally understand how these problems are done.

How is 4 < f(x) < 8 the same as |f(x)-6|< 2? f(x) in this case is 6. Therefore, how is 4 < 6 < 8 also represented in 0 < 2? Another question in the end; I understood when you subtracted 4 from all parts of the inequality, and resulted with -2< x- 4< 4. However, I got confused in the part when you said "since |-2|< |4|, that will be true as long as |x- 4|< |-2|= 2. " How do we know |x - 4| is less than |-2|? We only knew it was smaller than 4 and greater than -2.

I'm sorry for the additional hassle. I just want to get a grip on this.
• Sep 22nd 2010, 02:39 AM
HallsofIvy
Quote:

Originally Posted by FullBox
Thank you very much! I really appreciate your help and you did indeed provide the correct answer. However, there are still several things I do not understand and would really appreciate if you could clarify (I really want to understand and feel comfortable with this topic).

I completely understand how you restricted the function with the vertical and horizontal lines. Epsilon was 2, and so you restricted the function horizontally from y= 6 + ε = 8, and y = 6 - ε = 4. I also understand that the vertical lines restricted the graph from where the function intersected with the aforementioned horizontal lines.

However, in the end, I got a bit confused [it seems I got confused at the easiest part of the explanation]. I know this will come off as a stupid question, but I just want to totally understand how these problems are done.

How is 4 < f(x) < 8 the same as |f(x)-6|< 2? f(x) in this case is 6.

No, f(x) is NOT 6. x is a variable and so is f(x). You are mistaking "x", the variable, for " $x_0$, the value of that variable at a particular point.

Quote:

Therefore, how is 4 < 6 < 8 also represented in 0 < 2?
No, |f(x)- 6|< 2 is NOT the same as "0< 6< 2" (it is not even the same as 0< |f(x)- 6|< 2) nor is 4< f(x)< 8 the same as "4< 6< 8". Saying that |f(x)- 6|< 2 tells us that f(x) is some number that satisfies |f(x)- 6|< 2 which means, as I said before, that f(x) itself is some number between 6- 2= 4 and 6+ 2= 8. It is possible that f(x) is almost as low as 4, in which case f(x)- 6 is just slightly larger than -2 and so |f(x)- 6|< 2. It is possible that f(x) is almost at high as 8, in which case f(x)- 6 is slightly less than 2 and again |f(x)- 6|< 2.

Quote:

Another question in the end; I understood when you subtracted 4 from all parts of the inequality, and resulted with -2< x- 4< 4. However, I got confused in the part when you said "since |-2|< |4|, that will be true as long as |x- 4|< |-2|= 2. " How do we know |x - 4| is less than |-2|? We only knew it was smaller than 4 and greater than -2.
We don't. You are going the wrong way. I am saying that if |x- 4|< 2, then we have x- 4 between -2 and 2 and so certainly between -2 and 4. To be able to write |x- a|< b, I have to have a "symmetric" interval so I just "cut off" one end.

Quote:

I'm sorry for the additional hassle. I just want to get a grip on this.