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Homework Help: Squeeze Theorem/Limits Question. Answers don't match up.

  1. Jun 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate using the Squeeze Theorem.
    [itex] \lim_{x\rightarrow0} (2^{x} - 1) cos\frac{1}{x} [/itex]

    2. Relevant equations

    3. The attempt at a solution

    [itex] -1 ≤ cos\frac{1}{x} ≤ 1 \\
    1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\
    0 ≤ (2^{x} -1)cos\frac{1}{x} ≤ 0 \\
    ∴ \lim_{x\rightarrow0} (2^{x} - 1) cos\frac{1}{x} = 0 [/itex]

    That's the solution I got and the solution that's in my textbook. But when I checked on wolframalpha, it says the limit does not exist.

    Did I do something wrong and is my textbook wrong? or is it something conceptually that I don't understand about this? Is wolframalpha wrong?
    Thanks :)
  2. jcsd
  3. Jun 17, 2013 #2


    Staff: Mentor

    It's possible that you inadvertently entered something different into wolfram. Your answer is correct.
  4. Jun 18, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are correct that the limit is zero, but your reasoning has some flaws. The last line is clearly not correct, for it is equivalent to
    $$(2^x - 1)\cos\left(\frac{1}{x}\right) = 0$$
    which is not true for all ##x##. The previous line is correct, but only if ##2^x - 1 \geq 0##.

    It's somewhat easier to work with absolute values:
    $$\left|\cos\left(\frac{1}{x}\right)\right| \leq 1$$
    Now multiply both sides by ##|2^x - 1|## and see what you can conclude.
  5. Jun 19, 2013 #4

    for x < 0, this statement is false.
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