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anonymous

  • one year ago

Review: Limits Evaluate. Review: Limits Evaluate. \(\sf \Large lim_{x \rightarrow 0^+} \sqrt{x}e^{sin(\frac{\pi}{x})}\)

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  1. anonymous
    • one year ago
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    I assume that since \(\sf \large lim_{x \rightarrow 0^+} \sqrt{x}\) is zero, this limit will be equal to zero too. Because anything that we multiply with zero is zero and the limit law says that \(\sf lim_{x \rightarrow a} [f(x)g(x)]=lim_{x \rightarrow a} f(x) •lim_{x \rightarrow a} g(x) \) so for my question: \(\sf = lim_{x \rightarrow 0^+} \sqrt{x} • lim_{x \rightarrow 0^+} e^{sin(\frac{\pi}{x})}\\= 0 \ •\ lim_{x \rightarrow 0^+} e^{sin(\frac{\pi}{x})}\) i know that pi/x will be undefined and my reasoning is not that good, so yeah I need help with explaining it.

  2. ganeshie8
    • one year ago
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    are you allowed to use L'hopital ?

  3. Astrophysics
    • one year ago
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    I think squeeze theorem

  4. zepdrix
    • one year ago
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    ya squeeze seems nice for this one :)

  5. anonymous
    • one year ago
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    l'hopital's rule is not yet allowed to use in this problem

  6. Astrophysics
    • one year ago
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    Since \[- 1 \le \sin \left( \frac{ \pi }{ x } \right) \le 1\]

  7. zepdrix
    • one year ago
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    \[\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le1\]Exponentiate everything,\[\large\rm e^{-1}\le e^{\sin\left(\frac{\pi}{x}\right)}\le e^{1}\]Throw some of that square root magic into there,\[\large\rm \sqrt x e^{-1}\le \sqrt x e^{\sin\left(\frac{\pi}{x}\right)}\le \sqrt x e^{1}\]

  8. zepdrix
    • one year ago
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    And then let x approach 0, for the left and right most sides, ya? :o

  9. Astrophysics
    • one year ago
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    Looks good!

  10. anonymous
    • one year ago
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    oh okay,got it! :D Thank you so much!

  11. zepdrix
    • one year ago
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    yay team \c:/

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