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anonymous

  • 5 years ago

Prove the statement using the epsilon, sigma definition of limit. lim (7-3x)=-5 x--->4

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  1. anonymous
    • 5 years ago
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    I am struggling with these. :( Would be nice if some1 could show it.

  2. anonymous
    • 5 years ago
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    i have no idea

  3. anonymous
    • 5 years ago
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    you should start with something like this: for all epsilon (I will use E) >0 there exist a delta (D) such that D>|x-4| implies E>|f(x)-(-5)| Or something like this

  4. anonymous
    • 5 years ago
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    but what does that equation mean and how do you solve it

  5. anonymous
    • 5 years ago
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    good question :D I am looking at my notes now and trying to figure it out but not working well

  6. anonymous
    • 5 years ago
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    me too

  7. anonymous
    • 5 years ago
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    I am doing real analysis, are u too?

  8. anonymous
    • 5 years ago
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    no

  9. anonymous
    • 5 years ago
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    I'll be back in 10 min and give it another go

  10. anonymous
    • 5 years ago
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    So the definition is \[given any \epsilon >0 there \exists \delta >0 such that 0<\left| x-a \right|<\delta \implies \left| \left| f(x)-L \right| \right|<\epsilon\]

  11. anonymous
    • 5 years ago
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    a is 4 L is -5 here

  12. anonymous
    • 5 years ago
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    f(x)=7-3x

  13. anonymous
    • 5 years ago
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    ok so all you do is plug in the 4 for x which gives the answer as 7-12=-5, -5=-5 and that is it. -Source a friend who got an a in calc 3

  14. anonymous
    • 5 years ago
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    Yes that is fine, but it does not use epsilon or delta...

  15. anonymous
    • 5 years ago
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    I guess here you need to use this definition but I'm not sure how. :(

  16. anonymous
    • 5 years ago
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    i give up

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