Evaluate $$\lim_{n \rightarrow \infty} n \sin (2\pi e n!).$$
I wanna ask what's wrong with my method:
Define $C_n= n \cos (2\pi e n!)$ and $S_n=n \sin (2\pi e n!)$, then $C_n+iS_n=ne^{i2\pi en!}=n1^{en!}=n$, comparing the imaginary parts yeilds $S_n=0$. Therefore the limit equals $0$. Also, I am asking for a way to solve the problem, thank you.
Here is the error in your solution:
We are used to thinking of exponential functions as having a single well-defined value. However, if you admit complex numbers, then the expression $z^w$ can be interpreted infinitely many ways: there are infinitly many logarithms of $z$, all differing by multiples of $2\pi i$, and so $z^w$ might mean any of the quantities $e^{w(\log z + 2\pi i k)}$. Therefore saying that $1^{en!}=1$ is not valid: it might equal any of the quantities $e^{en!(0+2\pi ik)}$.
It is true that when we write $z^w$ when $z$ is a positive real number, we almost always mean one specific branch/incarnation of this function, namely the one that takes positive real values for $w$ positive and real. Under this convention, both expressions $e^{i2\pi en!}$ and $1^{en!}$ would be unambiguous; however, the usual laws of exponents are not valid when the exponents are complex. (In fact you've just proved that these laws are invalid, by proving that $n\sin(2\pi en!)=0$ for every positive integer $n$, which in particular "proves" that $e$ is half an integer!)
