Limit of dirichlet function (from DSP)

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SUMMARY

The limit of the Dirichlet function is evaluated as follows: \lim_{k->0}\frac{sin(\pi k)}{sin(\frac{\pi k}{N})} = N. The solution involves applying L'Hôpital's rule, leading to \lim_{k->0}N\frac{cos(\pi k)}{cos(\frac{\pi k}{N})}, which simplifies to N since cos(0) = 1. Additionally, utilizing the series expansion of the sine function provides an alternative method for evaluating this limit.

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Jyan
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How is this limit evaluated?

[tex]\lim_{k->0}\frac{sin(\pi k)}{sin(\frac{\pi k}{N})}[/tex]

I know that it is N, but I can't figure out how to evaluate it, l'hospital's rule doesn't seem to help.

I might solve it by the time I get a response, but figured no reason to not ask especially since I couldn't find much about it on Google.

Solved it, feel like an idiot:

[tex]\lim_{k->0}\frac{sin(\pi k)}{sin(\frac{\pi k}{N})}[/tex]

Using l'hospital's rule:

[tex]\lim_{k->0}N\frac{cos(\pi k)}{cos(\frac{\pi k}{N})}[/tex]

This is equal to N, since cos(0) = 1.
 
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