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amorable

  • 3 years ago

Squeeze Theorem Show that -|x| less than equal to x sin (1/x) less than equal to |x| The use the squeeze theorem to show that the limit as approaches 0 of x sin (1/x)=0 Can someone walk me through the squeeze theorem? I don't understand how to solve the problems.

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  1. LolWolf
    • 3 years ago
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    Well, to show both cases we simply have to show that \(|x|\ge xsin\left(\frac{1}{x}\right)\). Since we know, \(\forall x\in \mathbb{R}, |\sin(x)|\le1\), then, multiplying both sides by \(x\) gives us: \[ x|\sin(x)|\le x\implies x\sin(x)\le x \]And thus, since \(|x|\ge x\) \[ x\sin(x)\le|x| \]We know: \[ -|x|\le x \]Which means that (given some more, similar work to the above): \[ x\sin(x)\ge-|x| \]Therefore: \[ -|x|\le x\sin(x) \le |x| \]And, we are done.

  2. LolWolf
    • 3 years ago
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    For the second part, we know: \[ \lim_{x\to 0}-|x|=0\\ \lim_{x \to 0}|x|=0 \]And, since: \[ -|x|\le x\sin\left(\frac{1}{x}\right) \le |x| \]We then have that, by the squeeze theorem. \[ \lim_{x\to 0}\left(x\sin\left(\frac{1}{x}\right)\right)=0 \] And I just realized I made a mistake on the last post. Everytime I mention \(x\sin(x)\) I actually mean \(x\sin\left(\frac{1}{x}\right)\). My bad.

  3. amorable
    • 3 years ago
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    Thank you! The way you put it makes it easier to understand than the book. Does this work the same for cosine and secant limits?

  4. LolWolf
    • 3 years ago
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    Yes for cosine, as for secant, it's slightly different, but try it yourself, it's more fun, that way. And sure thing.

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