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anonymous
 5 years ago
What is the limit of (x Sin (1/x)) as x becomes positive infinite?
anonymous
 5 years ago
What is the limit of (x Sin (1/x)) as x becomes positive infinite?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i believe the limit is 0 because as x increases, 1/x decreases to where it is approaching zero and the sin (0) is 0. and anything multiplied by 0 is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use a substitution u=1/x, then the limit will be \[\lim_{u \rightarrow 0}{\sin x \over x}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry \[\lim_{u \rightarrow 0} {\sin u \over u}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0limits is zero , its seen by pinching theorm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anwar..i think you might have to use L'Hopital's Rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think L'hopital's rule can work here. I still believe it's 1 by the substitution I used.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0check it, sub in a big number'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the larger you're number gets, the closer you are to zero because it is a fraction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@thamir: When substituting \(u=1/x\), \(\sin (1/x)= \sin u\) and \(x=1/u\). Also the approaching point will be zero instead of infinity since \(1/\infty \rightarrow 0\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01<sin(1/x) < 1 , this is known , multiply through by x ( it is positive because we are considering x>+infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, thats not going to work :

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0something similar lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@elecengineer: When you substitute a large number the x goes to \(\infty \) and the sin(1/x) goes to \(0\). And \( \infty \times 0 \ne0\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0L'Hopitals Rule does work AnwarA and you are right the answer is 1. lim sin(1/x)/(1/x) =lim [(1/x^2)cos(1/x)]/(1/x^2) =lim cos(1/x) =cos(0)=1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \inf}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}\] let u=1/x x>inf, u>0 so the limit is 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks AnwarA for solving it. Thanks chris777 for teaching me the L'Hopitals Rule. And last but not least thanks to elecengineer for trying to solve using the squeeze theorem. Now, I need to get the correct answer which is 1 by using the squeeze theorem. When trying I reached the same point where elecengineer reached. How can we solve it using the squeeze theorem?
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