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Thread: prove limit by definition

  1. #1
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    prove limit by definition

    prove that $\displaystyle lim_{x\rightarrow2}\frac{x}{x+1}\neq1$ by epsilon delta
    defintion.
    we need to prove that their is $\displaystyle \varepsilon>0$ so for every $\displaystyle \delta>0$
    their is x $\displaystyle |x-2|<\delta$ and $\displaystyle |\frac{x}{x+1}-1|\geq\varepsilon$
    so we are free to choose our epsilon and our x
    i choose $\displaystyle \varepsilon=2$
    x cannot be equal to -1 because of the denominator.
    and we need to choose x for which $\displaystyle |\frac{x}{x+1}-1|\geq2$
    $\displaystyle |\frac{x-x-1}{x+1}|\geq2$ -> $\displaystyle |\frac{1}{x+1}|\geq2$ -> $\displaystyle \frac{1}{2}\geq|x+1|$
    -> $\displaystyle -\frac{1}{2}\leq x+1\leq\frac{1}{2}$ -> $\displaystyle -\frac{3}{2}\leq x\leq-\frac{1}{2}$
    but x=-1 there so we have a problem.
    where is the problem in my way???


    the prof solved it by $\displaystyle \varepsilon=1/6$
    $\displaystyle 2>x>max\{2-\delta/2,1\}$
    cant understand why he choose such x
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  2. #2
    Grand Panjandrum
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    Re: prove limit by definition

    Quote Originally Posted by transgalactic View Post
    prove that $\displaystyle lim_{x\rightarrow2}\frac{x}{x+1}\neq1$ by epsilon delta



    defintion.
    we need to prove that their is $\displaystyle \varepsilon>0$ so for every $\displaystyle \delta>0$
    their is x $\displaystyle |x-2|<\delta$ and $\displaystyle |\frac{x}{x+1}-1|\geq\varepsilon$
    so we are free to choose our epsilon and our x
    i choose $\displaystyle \varepsilon=2$
    x cannot be equal to -1 because of the denominator.
    and we need to choose x for which $\displaystyle |\frac{x}{x+1}-1|\geq2$
    $\displaystyle |\frac{x-x-1}{x+1}|\geq2$ -> $\displaystyle |\frac{1}{x+1}|\geq2$ -> $\displaystyle \frac{1}{2}\geq|x+1|$
    -> $\displaystyle -\frac{1}{2}\leq x+1\leq\frac{1}{2}$ -> $\displaystyle -\frac{3}{2}\leq x\leq-\frac{1}{2}$
    but x=-1 there so we have a problem.
    where is the problem in my way???


    the prof solved it by $\displaystyle \varepsilon=1/6$
    $\displaystyle 2>x>max\{2-\delta/2,1\}$
    cant understand why he choose such x
    Since the actual limit is 2/3, you will always be able to show that for x close enough to 2 that |x/(x+1)-1|>0.3, so you need to choose epsilon greater than zero but less than 0.3 (well actually any thing less than 1/3 will do). Then show that for all delta>0 there is always some x such that |x-1|<delta and |x/(x+1) - 1|>epsilon.

    CB
    Last edited by CaptainBlack; Aug 9th 2011 at 10:29 AM. Reason: correct typo, and adjust the odd value
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  3. #3
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    Re: prove limit by definition

    what is the problem of my way.
    ??
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  4. #4
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    Re: prove limit by definition

    Consider $\displaystyle {\delta}$ <_1. Then if x satisfy's 0< |x-2|< $\displaystyle {\delta}$, F(x)<3/4. Hence |F(x)-1|>1/4. Choosing $\displaystyle {\epsilon}$<_1/4 completes the proof.
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  5. #5
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    Re: prove limit by definition

    i cant because its for every delta

    and can you tell me where am i wrong
    ?
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  6. #6
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    Re: prove limit by definition

    For any delta>1, x values in (1,3) will still have to satisfy |F(x)-1|< epsilon.
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  7. #7
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    Re: prove limit by definition

    It's not true that you are totally free to choose any ε that you like.

    As CaptainBlack told you, the actual limit is 2/3.

    |1 - 2/3| = 1/3. So, if you use an ε ≥ 1/3, it is possible to find a δ > 0 such that for all x for which 0 < | x - 2 | < δ it follows that | x/(x+1) - 1 | < ε .

    That's why CB said you need to choose ε < 1/3 .
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