<s>[INST] Value of the limit without (or with, but giving rigorous arguments) using the Taylor expansion of sin I'm trying to evaluate the limit as $N\to \infty.$
$$\frac{  \left(\dfrac{\sin \frac{1}{N}} {\frac{1}{N}}\right)^{N}   -1 }{\frac{1}{N}}.$$
Note first that, using L'Hôpital, one can easily show that the numerator goes to $0.$
Using the Taylor series expansion for $sin$, the value of the actual limit seems to be  $-\frac{1}{6}$. But I'm not fully sure how to justify the infinite series in the numerator is $-\frac{1}{6N}+O(\frac{1}{N^2})$. You could either justify that, if I was right, or may be use some other method to tell me what the limit is?
Many thanks in advance!
 [/INST] Just develop carefully the Taylor expansion.
$$\sin \frac{1}{N}=\frac{1}{N}-\frac{1}{6N^3}+O(\frac{1}{N^5})$$
Then
$$\left(\frac{\sin \frac{1}{N}}{\frac{1}{N}}\right)^N=\exp \left[N \log\left( 1-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left[N\left(-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left(-\frac{1}{6N}+O(\frac{1}{N^3})\right)=1-\frac{1}{6N}+O(\frac{1}{N^2})$$
And
$$\frac{\left(\frac{\sin \frac{1}{N}}{\frac{1}{N}}\right)^N-1}{\frac{1}{N}}
=N\left(-\frac{1}{6N}+O(\frac{1}{N^2})\right)=-\frac{1}{6}+O(\frac{1}{N})$$
Hence the limit as $N\to\infty$ is $-\frac{1}{6}$.
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