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Prove that the lim (x->1) x^2 + 2 is NOT equal to 2.999?

  1. Oct 19, 2004 #1
    How can we prove that the lim (x->1) x^2 + 2 is NOT equal to 2.999? (example I made up right now) At the end of each proof we find a relation between epsilon and delta. What does it mean?
     
  2. jcsd
  3. Oct 19, 2004 #2
    You cant take back the 2 as a constant, and the limit of x² will be 1
     
  4. Oct 19, 2004 #3

    Galileo

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    Since
    [tex]\lim_{x\rightarrow 1}x^2+2=2.999 \iff \lim_{x\rightarrow 1}x^2=0.999[/tex]
    we can work with the righthanded expression. (it's easier).
    We'll have to find an [itex]\epsilon>0[/itex] such that for any [itex]\delta>0[/itex] we have [itex]|x-1|<\delta [/itex] AND [itex]|x^2-0.999|>\epsilon[/itex]

    EDIT: Removed 'proof', because of errors.
     
    Last edited: Oct 19, 2004
  5. Oct 19, 2004 #4
    my friend it's real simple; as the 2 won't be effected by the limit, so only part affected by limit is x^2 when x->1

    although u are write it might go to 2.999 when it approach from left, but u need to consider another fact when limit approach from right than it will be 3.00001 that's how we prove that because u don't specify as limit apporaches from right or left
     
  6. Oct 20, 2004 #5
    Proving that [tex] \lim_{x\rightarrow a} f(x) = b [/tex] means that it should be possible, by choosing a [tex] \delta [/tex] , to make the make the difference in values of f(x) and b smaller than any positive number [tex]\epsilon [/tex], for all the values of [tex]|x-a|[/tex] less than [tex] \delta [/tex].

    For 3, you can show this to be true. But for 2.999, although you can choose a number for which the difference is smaller than any positive number, but this will not be true for all the [tex] |x-a|[/tex] smaller than that chosen number.

    That is the reason why 2.999 isn't the limit of the function.


    spacetime
    www.geocities.com/physics_all/index.html
     
  7. Oct 21, 2004 #6
    These replies make think, once again, what does it mean to be "close"?
    Can't we find a relation between epsilon and delta by doing the operations abs(x^2 + 2 - 2.999) < epsilon and abs(x-1) < delta?
    I'm really stuck with this issue for two weeks and think I will not understand the concept of the formal definition of a limit. I've tried nearly anything on the web :(
     
  8. Oct 21, 2004 #7

    arildno

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    Sure, here's how:
    [tex]|x^{2}+2-2.9991|=|(x-1)(x+1)+r|,r=0.0009[/tex]
    1) Assume x>1:
    Then r+(x-1)(x+1)>r>0, so:
    [tex]|(x-1)(x+1)+r|>r[/tex].
    Choose in this case, [tex]\epsilon<r[/tex]
    2)Assume x<1.
    Then, you can find [tex]\delta[/tex], so that [tex](1-x)(x+1)<\frac{r}{2}[/tex]
    Hence, by the triangle inequality, we have:
    [tex]|x^{2}+2-2.9991|=|(x-1)(x+1)+r|\geq(r-(1-x)(x+1))\geq(r-\frac{r}{2})=\frac{r}{2}[/tex]

    Hence, the choice [tex]\epsilon<\frac{r}{2}[/tex] is always out of reach, so 2.9991 cannot be the limit.
     
    Last edited: Oct 21, 2004
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