anonymous
  • anonymous
Can someone explain to me the precise definition of a limit?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Liv1234
  • Liv1234
"In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value. Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives, and integrals." ~Wikipedia.
Liv1234
  • Liv1234
The definition of a limit should be in your math book, though.
freckles
  • freckles
|dw:1449967354817:dw| let delta and epsilon me greater than 0 the smaller you choose delta the smaller epsilon is and the closer you are to your limit L

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freckles
  • freckles
|dw:1449967500076:dw|
anonymous
  • anonymous
ok, I understand that. but when it comes to proving limits. for example. Prove \[\lim_{x \rightarrow 3} (4x-5)=7\] how would you go about proving this?
freckles
  • freckles
oh you have to pick a delta such that |f(x)-L|
freckles
  • freckles
\[\text{ example } \\ \lim_{x \rightarrow -2} (2x-1)=-5 \\ |f(x)-L| < \epsilon \\ |2x-1-(-5)| < \epsilon \\ \\ |2x-4| < \epsilon \\ \text{ you basically want \to solve this for } |x-a| \text{ or and this case } |x-(-2)| \text{ or } |x+2| \\ \\ |2||x-2|< \epsilon \\ 2 |x-2| < \epsilon \\ |x-2| < \frac{\epsilon}{2} \\ \text{ so you would want to pick } \delta \text{ to be } \frac{\epsilon }{2}\]
freckles
  • freckles
I typo causing me to mess up later
freckles
  • freckles
-1+5 is +4 not -4
freckles
  • freckles
it should have been |x+2| all the way down
anonymous
  • anonymous
ok so delta= epsilon/2
anonymous
  • anonymous
but then what is epsilon?
freckles
  • freckles
\[\text{ example } \\ \lim_{x \rightarrow -2} (2x-1)=-5 \\ |f(x)-L| < \epsilon \\ |2x-1-(-5)| < \epsilon \\ \\ |2x+4| < \epsilon \\ \text{ you basically want to solve this for } |x-a| \text{ or and this case } |x-(-2)| \text{ or } |x+2| \\ \\ |2||x+2|< \epsilon \\ 2 |x+2| < \epsilon \\ |x+2| < \frac{\epsilon}{2} \\ \text{ so you would want to pick } \delta \text{ to be } \frac{\epsilon }{2} \] \[\text{ Proof } \\ \text{ Let } \epsilon >0. \text{ Choose } \delta =\frac{\epsilon }{2} \\ \text{ and so } |f(x)-L|=|2x-1-(-5)|=2|x+2|<2 \frac{\epsilon}{2}=\epsilon \\ \\ \text{ EOD}\] the definition is just wanting us to choose a delta such that the |f(x)-L|
freckles
  • freckles
this is called an existential proof where you get to choose delta the word there exist a delta is key into knowing that
anonymous
  • anonymous
oh ok, so we just do everything in terms of epsilon
freckles
  • freckles
I think I will say yes you don't get to choose epsilon you have to find a delta that works for every epsilon
anonymous
  • anonymous
and then if |f(x)-L|=epsilon the limit is true/
freckles
  • freckles
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jim_thompson5910
  • jim_thompson5910
The idea is that epsilon is some small number. You can make it as small as you want and the proof @freckles showed is that no matter what value epsilon is, there is a delta that corresponds to it. So that effectively proves what this page is saying https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html
anonymous
  • anonymous
Okay I think I got it

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