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Inequalities/limits question

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm proving the limit of an equation with the epsilon-N notation for negative infinity.

    Here is the equation that I'm trying to prove.
    lim [tex]\frac{1}{x} = 0[/tex]
    [tex]x \rightarrow -\infty[/tex]

    I get stuck at the inequality point.

    So, let [tex]\epsilon[/tex]>0, N<0 (N is negative) such that

    [tex]\left|\frac{1}{x} - 0\right| < \epsilon[/tex] whenever x<N

    So, [tex]x > \frac{1}{\epsilon}[/tex]

    But we want N to be negative, so add a negative sign

    [tex]x > -\frac{1}{\epsilon}[/tex]

    Now this is where I get confused.

    I have that [tex]x > -\frac{1}{\epsilon}[/tex], yet I want x<N. If N is [tex]-\frac{1}{\epsilon}[/tex], then x not less than N. If x is not less than N, then wouldn't it be wrong to use [tex]-\frac{1}{\epsilon}[/tex] for N, because our x<N has not been satisfied. Or have I got an inequality wrong somewhere?

  2. jcsd
  3. Nov 13, 2008 #2


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    Staff Emeritus
    Science Advisor

    No. You can't just "add a negative sign". Since we know that know that x is negative, |1/x|= -1/x and [itex]|1/x|< \epsilon[/itex] is immediately [itex]-1/x< \epsilon[/itex] and, multiplying both sides by the negative number [itex]x/\epsilon[/itex], [itex]-1/\epsilon> x[/itex]

    Yes, you have the inequality wrong. "Add a negative" is not an algebraic operation. You have to be more careful with your manipulation of the absolute value.
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