Calculating Limits: Refresh Your Memory

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SUMMARY

The limit of the expression \(\lim_{n\to\infty} \frac{\sin(nt)}{n}=0\) for \(t \in [0,1]\) is established through the properties of bounded functions and the behavior of sequences. Since the sine function is bounded by -1 and 1, the fraction \(\frac{\sin(nt)}{n}\) approaches zero as \(n\) increases indefinitely. This conclusion is derived from the fact that the numerator remains bounded while the denominator grows without bound.

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  • Understanding of limits in calculus
  • Familiarity with the properties of the sine function
  • Basic knowledge of sequences and series
  • Concept of bounded functions
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  • Review the definition of limits in calculus
  • Study the properties of bounded functions and their implications
  • Explore examples of limits involving trigonometric functions
  • Learn about sequences and their convergence criteria
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Students of calculus, educators teaching limit concepts, and anyone looking to refresh their understanding of trigonometric limits.

Ted123
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Show that, with [itex]t\in [0,1][/itex] : [tex]\lim_{n\to\infty} \frac{\sin(nt)}{n}=0[/tex]

This is easy but I've forgotten how to calculate limits. Can anyone jog my memory?
 
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Not sure, but how can the limit be something else when |sin| <=1 and n goes to infinity?
 

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