Question

Anyone up for more limits with epsilon and delta?

*Def 2 says that as x approaches a, the limit of f(x) =L

Answer 1

Ummm can you check one for me? >.<
I wanna see if I'm doing this correctly or not.

For epsilon = 0.5 my answer is 1/8

Answer 2

CORRECT!

Answer 3

Can you show your work please? I don't want just the answer :) you work to hard for me to just take it.

Answer 4

Yes yes, I just didn't want to go explaining something if I was doing it way wrong >.<

Answer 5

For all \(\Large\rm \epsilon\gt0\) there exists a \(\Large\rm \delta\gt0\) such that,
if,\[\Large\rm 0\lt|x-1|\lt \delta\]then,\[\Large\rm |4x-3-1|\lt \epsilon\]\[\Large\rm |4x-4|\lt \epsilon\]\[\Large\rm |x-1|\lt \frac{\epsilon}{4}\]Ooo my next step was a little sloppy.
I plugged delta in for the |x-1|.
I think I read one of my inequalities backwards.
I'll have to rethink how I justified that.
\[\Large\rm \delta <\frac{\epsilon}{4}\]

Answer 6

Editorial note: zepdrix wrote\[\Large\rm |4x-3-1|\lt \epsilon\]after having replaced f(x) in Definition 2 with the given function, 4x-3.

Answer 7

Suggestion: Copy down Def 2, first in the most general form, and then second, replacing f(x) with the given function.

Note that \[\Large\rm |4x-3-1|\lt \epsilon\]reduces to \[|4x-4|<\]...which in turn simplifies to \[4|x-1|<\epsilon\]

Answer 8

Again referring to Def #2, can you now determine the value of little delta?

Answer 9

...considering that epsilon = 0.5, \[4|x-1|<0.5\]

Answer 10

Hint: divide both sides of this inequality by 4. Why?

Answer 11

*web site loses connection, bear with me please*

Answer 12

\[\left| x-1 \right|<\frac{ .5 }{4 }\]

Answer 13

Then add 1...

Answer 14

Sorry, I like writing better :)

Answer 15

Whoooo. Mistake on my end.

Answer 16

Well be careful.
if you wanted to add 1, make sure you deal with the absolute bars first.\[\Large\rm \left| x-1 \right|<\frac{ .5 }{4 }\]\[\Large\rm -\frac{ .5 }{4 }\lt x-1 <\frac{ .5 }{4 }\]

Answer 17

But then you can, right?

Answer 18

ya :D