geerky42
  • geerky42
How can I prove that \(\displaystyle \lim_{x\to25}\sqrt x = 5\) using Delta-Epsilon?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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geerky42
  • geerky42
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?
DanJS
  • DanJS
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
geerky42
  • geerky42
Yeah, and I am having trouble finding this function.

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geerky42
  • geerky42
NOTE: I am self teaching
DanJS
  • DanJS
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 \[0 < \left| x-a \right|<\delta\] \[\left| f(x) - L \right| < \epsilon\]
DanJS
  • DanJS
it sorta limits you to a box around the point
tkhunny
  • tkhunny
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".
DanJS
  • DanJS
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)
DanJS
  • DanJS
sorta a box around the point in question at the center of it sides a-delta and a+delta, L-epsilon and L+epsilon
DanJS
  • DanJS
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
DanJS
  • DanJS
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
geerky42
  • geerky42
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?
geerky42
  • geerky42
I am just confused about relationship
geerky42
  • geerky42
I have to PROVE the limit, you know.
DanJS
  • DanJS
here is the same function , just a different 'a' value to use and L
DanJS
  • DanJS
https://www.youtube.com/watch?v=JEOwCF9rXwo
tkhunny
  • tkhunny
Why would you ever maintain \(\epsilon^{2}\)? That's way too small to worry about.

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