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How to prove a limit is correct

  1. Nov 2, 2004 #1
    The rigorous definition given in my calculus book is as follows
    [tex]\lim_{x\to{a}}f(x) = L[/tex] if given any number [tex]\epsilon > 0[/tex] we can find a number [tex]\delta > 0[/tex] such that [tex]| f(x) - L | < \epsilon[/tex] if [tex] 0 < | x - a | < \delta[/tex]

    How can I use that to prove [tex]\lim_{x\to{49}} x^{\frac{1}{2}} = 7[/tex]

    This is what I did
    [tex]| f(x) - L | < \epsilon[/tex]
    [tex]| x^{\frac{1}{2}} - 7 | < \epsilon[/tex]
    [tex]| x - 7x^{\frac{1}{2}} + 49 | < \epsilon^2[/tex]

    That's where I get stuck and don't know where to go. This is the first time I've encountered any epsilons and deltas so its all confusing. Thanks for any help
  2. jcsd
  3. Nov 2, 2004 #2


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    Hi, it looks like you're are assuming from the start that [tex]| f(x) - L | < \epsilon[/tex], this isn't the way to go. Does your textbook have any examples that you've looked over to get the general feel? If this is your first time seeing an epsilon delta limit I'd suggest you try looking at a few examples first. Then try some easier examples than the one you've posted, like show

    [tex]\lim_{x\rightarrow 3}x+5=8[/tex]
    [tex]\lim_{x\rightarrow -1}x^2-2x+3=6[/tex]

    These will be more straightforward than your question with the root, yet will be handled similarily. Do you think you could do the above limits? If not, I'd be willing to work through them for you, before you try to tackle the question you've given.
  4. Nov 3, 2004 #3


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    No, starting from |f(x)-L|< ε and then showing that you can solve for δ is perfectly reasonable. (It's sometimes called "synthetic proof": start from what you want to show and derive a statement that is obviously true. As long as eveery step is "reversible", the real proof would be starting from the obviously true statement and working back to what you wanted to show.)

    In this case, you want to show that [itex]\lim_{x\to{49}} x^{\frac{1}{2}} = 7[/itex] so you want to show that "for any ε> 0, there exist δ> 0 so that if |x-49|< δ then [itex]|x^{\frac{1}{2}}-7|< \epsilon[/itex]".

    Okay, start from [itex]|x^{\frac{1}{2}}-7|< \epsilon[/itex].
    You apparently squared both sides. I wouldn't do that. What I would do is use
    x2- a2= (x-a)(x+a). Multiply both sides of the inequality by
    [itex]x^{\frac{1}{2}}+ 7[/itex] and see what you get.
    Last edited by a moderator: Nov 3, 2004
  5. Nov 3, 2004 #4
    you wind up with [tex]| x - 49 | < ( \epsilon \times ( x^{1/2} + 7) )[/tex]

    so I'm assuming that with that last statement, I've found a [tex]\delta[/tex] that works ([tex]\delta = \epsilon \times ( x^{1/2} + 7 )[/tex] ?
  6. Nov 3, 2004 #5

    matt grime

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    well, you now need to choose a delta to work backwards. Ths working should have told you approximately what form the delta is (it isn't correct right now because of the x in it.)
  7. Nov 3, 2004 #6
    This comes as a complete surprise for me. I always thought that you had to start from a statement which doesn't already assume what you're trying to prove and then try to prove. It is perfectly acceptable to me but I've come to believe that perhaps the 'A' Level markers won't accept it, would they?
  8. Nov 3, 2004 #7

    matt grime

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    Notice that it is required that all deductions are if and only if statements, Ethereal. It is useful and instructive to see what happens working backwards to see how the proof should run. As long as you say at the end, hence we see doing blah leads us to the desired conclusion. However, A level doesn't use things that have proofs in this sense really. It is more usual to "show some result" or "find the component of forces in some direction is" and these are always easy to do directly.

    In general there is a way to answer every analysis question, in essence, of this type: suppose that |x-y| < d without saying what d is, then calculate |f(x) - f(y)| and see what happens, in terms of d, and then saying "ah, given e, we can pick d to make it happen"

    eg, show x^2 is continuous on the interval [0,1]

    if |x-y|<d, then |x^2-y^2| = |x-y||x+y| < 2d, so given e pick d such that d<e/2 and we are done.
  9. Nov 3, 2004 #8


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    I didn't mean to imply that you couldn't produce a valid proof this way, only that if you are new to epsilon-delta proofs and finding them confusing that this might be a bad idea. It's just my experience that many students who attempt a synthetic proof usually end up drowning.
  10. Nov 3, 2004 #9
    Just to intercede...

    My analysis professor said we should always start with some "scrapwork" working with the |f(x) - L| < [epsilon] inequality, because, quite frankly, what else could you work with? :P
  11. Nov 4, 2004 #10
    I was referring more to the "prove this trigo identity" type questions, where by I was wondering if you could simply factor out the RHS from the LHS and show how what's left reduces to 1.
  12. Nov 6, 2004 #11


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    you are trying to prove that for any number close enough to 49, its square root is close to 7. so the smart way to go is to take any old number close to 49 and try to see how close its square root is to 7.

    but i do not see how to do that, so how about working backwards and noticing that squaring is a strictly increasing function.

    i.e. take numbers clsoe to 7 and see how big their squares are.

    so just square 7+a and see what you get.

    well you get 49 + 14a + a^2. So you are trying to prove that if a is small enough, then this is as close as you like to 49.

    can you show that? i.e. can you make 14a + a^2 small, by making a small?

    ah phooey, i know this is trivial. let e be given and look at (7-e)^2 and (7+e)^2.

    Then we have (7-e)^2 < 49 < (7+e)^2. Hence since taking square root is also strictly monotone, the square root of any number between (7-e)^2 and (7+e)^2 will be between 7-e and 7+e. Thus by the definition of limit, just take d to be the smaller of the two distances, 49 - (7-e)^2 or 49 - (7+e)^2.

    I.e. then any number closer to 49 than that d, will be between (7-e)^2 and (7+e)^2. hence its square root woill be between 7-e and 7+e.

    This based on the trick of proving that the inverse of a continuous strictly monotone functrion is also continuous.
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