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Math Help - [SOLVED] finding a limit

  1. #1
    Junior Member
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    Jul 2009
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    [SOLVED] finding a limit

    \lim_{x \to 0} (e^x-x-1) \cos^2 \frac{2}{x} + x + 1

    I know that I need to use the squeeze theorem, and I have an explanation of how to solve the problem, but I don't understand one of the steps, how does:

    0 \leq \cos^2 \frac{2}{x} \leq 1 \Rightarrow
    x+1 \leq (e^x-x-1)\cos^2 \frac{2}{x}+x+1 \leq e^x ?
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  2. #2
    MHF Contributor
    Joined
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    Talking

    The cosine varies between -1 and 1, so the square, being non-negative, varies between 0 and 1. Now do the plug-n-chug:

    If you take the lower limit, then (exponential stuff)*(cosine stuff) is bounded below by (exponential stuff)*(zero), or zero.

    If you take the upper limit, then (exponential stuff)*cosine stuff) is bounded above by (exponential stuff)*(one), or one.

    If you're dealing with the lower limit, the expression simplifies as 0 + x + 1 = x + 1.

    If you're dealing with the upper limit, the expression simplified as (e^x - x - 1) + x + 1 = e^x + x - x + 1 - 1 = e^x.
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