Calculating Limits: Refresh Your Memory

Thread starter Ted123
Start date Nov 16, 2011
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Limits

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The limit of the expression lim (n→∞) (sin(nt)/n) equals 0 for t in the interval [0,1]. This is because the sine function is bounded between -1 and 1, meaning that sin(nt) will not exceed these values regardless of n. As n approaches infinity, the denominator grows without bound, causing the overall fraction to approach zero. The reasoning hinges on the fact that the numerator remains bounded while the denominator increases indefinitely. Thus, the limit confirms that the expression converges to zero.

Nov 16, 2011

Ted123

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

This is easy but I've forgotten how to calculate limits. Can anyone jog my memory?

Last edited: Nov 16, 2011

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Nov 16, 2011

MaxManus

Not sure, but how can the limit be something else when |sin| <=1 and n goes to infinity?

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