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Squeeze theorem

  1. Oct 31, 2004 #1
    Ok, this is kind of embarrassing. I'm halfway through calc II and while the squeeze theorem makes sense to me when I read it, I don't see how it's applied. I feel like I must be missing something very fundamental for this to not make sense.

    [tex]\lim_{x \rightarrow \infty}(cosx)/x = 0 [/tex]


    What are the 2 functions that are squeezing the 3rd? Once you know the 2 functions do just evaluate one of them at the limit?

    Is it as simple as just splitting up the function? [tex](1/x)(cos x)[/tex] and then evaluating [tex]1/x[/tex] at infinity. That would certainly yield 0.
     
  2. jcsd
  3. Oct 31, 2004 #2

    Hurkyl

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    Usually, you use the squeeze theorem by identifing a piece of your function that you can bound.

    Let me do an example in a slightly silly way:

    [tex]
    \lim_{x \rightarrow \infty} \frac{x^2 + \ln x}{x^2 + 1}
    [/tex]

    Now, when x is large, we have the bound [itex]0 < \ln x < x[/itex]. With a little work, we can prove:

    [tex]
    \frac{x^2}{x^2+1} < \frac{x^2 + \ln x}{x^2 + 1} < \frac{x^2 + x}{x^2+1}
    [/tex]

    for large x. (such as x > 1)

    We can evaluate the limits on the left and right hand sides: they both come out to 1. So, the original limit must also be 1!


    That's how the squeeze theorem is usually used: you identify a piece of the function that you can bound, then substitute in the bounds to get an inequality to which you can apply the squeeze theorem.
     
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