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anonymous
 one year ago
Review: Limits
Evaluate.
Review: Limits
Evaluate.
\(\sf \Large lim_{x \rightarrow 0^+} \sqrt{x}e^{sin(\frac{\pi}{x})}\)
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0are you allowed to use L'hopital ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2I think squeeze theorem

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5ya squeeze seems nice for this one :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0l'hopital's rule is not yet allowed to use in this problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Since \[ 1 \le \sin \left( \frac{ \pi }{ x } \right) \le 1\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm 1\le \sin\left(\frac{\pi}{x}\right)\le1\]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,\[\large\rm \sqrt x e^{1}\le \sqrt x e^{\sin\left(\frac{\pi}{x}\right)}\le \sqrt x e^{1}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5And then let x approach 0, for the left and right most sides, ya? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay,got it! :D Thank you so much!
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