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Prove a limit using inequalities

  1. Mar 18, 2012 #1
    Using the inequalities: [tex]\sin x \leq x \leq \tan x[/tex] valid in a zero range, prove that:

    [tex]\displaystyle\lim_{x \to{0}}{\frac{x}{\sin x}}= 1[/tex]

    Thank you!!
  2. jcsd
  3. Mar 18, 2012 #2
    I'm rather unsure about this, but I'm deeply interested in the answer. I have attempted a solution that may or may not be correct.

    What we're trying to prove is that for every epsilon > 0, there is some delta > 0 such that for all x, |x- 0| < delta and |x/sinx - 1| < epsilon. Because of this, we can assign a delta a particular value in terms of the epsilon.

    if sinx ≤ x ≤ tanx, then (1/sinx) ≥ (1/x) ≥ (1/tanx)

    Factor |(x/sinx) - 1| into |x| * |(1/sinx - 1/x)|

    |(1/sinx - 1/x)| ≤ |1/sinx| + |1/x| < |1/sinx| + |1/sinx| by a manipulation of our given inequality = |2/sinx|

    Let δ = min(1, ε/|2/sinx|)
    from |x| < δ
    we have |x| < ε/|2/sinx|
    |x|*|(1/sinx + 1/x)| < ε/|2/sinx|*|(1/sinx + 1/x)|
    |x/sinx - 1| < ε/|2/sinx|* |2/sinx| by inequality above
    |x/sinx - 1| < ε
    Last edited: Mar 18, 2012
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