Question

How can I prove that \(\displaystyle \lim_{x\to25}\sqrt x = 5\) using Delta-Epsilon?

Answer 1

While working backward, I got \(\epsilon^2-10\epsilon < x-25 < \epsilon^2+10\epsilon\)

I am not sure how to set relationship between \(\epsilon\) and \(\delta\) such that provided \(\epsilon>0\), we have \(\delta > 0\).

I can have \(\min(10\epsilon-\epsilon^2, \epsilon^2+10\epsilon)\), but it works only for \(\epsilon < 10\).

\(\delta\) should be greater than \(0\) for all \(\epsilon \in\mathbb R^+\), right?

Answer 2

the overall thing is -

however close you want to be to this limit value L, epsilon away from it

there will be a value for how close to that value , 'a', delta away from it


so there is a function that relates delta to epsilon

Answer 3

Yeah, and I am having trouble finding this function.

Answer 4

NOTE: I am self teaching

Answer 5

IF you are within epsilon of the limit value (say any distance you want for epsilon > 0

Then - there is a delta >0, such that

\[0 < \left| x-a \right|<\delta\]
\[\left| f(x) - L \right| < \epsilon\]

Answer 6

it sorta limits you to a box around the point

Answer 7

If YOU promise to wander no farther away from \(x_{0}\) than \(\delta\),
The FUNCTION promises to wander away from \(f(x_{0})\) no farther than \(\epsilon\).
There is not necessarily a single function that will show it to be the case. There are consensus techniques that lead to what might be called, "most instructive" or "most naturally motivated".

Answer 8

the proof part, is to show that you can move as close as you want to the 'a' value for the limit, abs(x-a)<delta

you can always show the value for the limit is less than the value f(x) - L < epsilon, that will become arbitrarily close to the limit as x approaches 'a'

Answer 9

sorta a box around the point in question at the center of it

sides a-delta and a+delta, L-epsilon and L+epsilon

Answer 10

The problem here is saying...

The limit exists If and Only Iff , confining the X value to delta units of 25
will inevitably confine sqrt(x) to epsilon units of 5

Answer 11

i watched this a little while ago, and it was pretty straight forward i think...
https://www.youtube.com/watch?v=G0ax2x2_Em0

couple examples follow also in others

Answer 12

Ok I understand what Delta-Epsilon is, I have trouble setting relationship,

Is \(\delta ~\stackrel{\text{set}}{=}~\begin{cases} 10\epsilon-\epsilon^2 & \mbox{if } \epsilon \le 9 \\ 9 & \mbox{if } \epsilon >9 \end{cases}\) good?

Answer 13

I am just confused about relationship

Answer 14

I have to PROVE the limit, you know.

Answer 15

here is the same function , just a different 'a' value to use and L

Answer 16

https://www.youtube.com/watch?v=JEOwCF9rXwo

Answer 17

Why would you ever maintain \(\epsilon^{2}\)? That's way too small to worry about.