Artificial intelligent assistant

Dominated convergence theorem on $e^{ix}$ I am considering first: $$\lim_{n \to 0} \int_{0}^{\pi} e^{ine^{ix}} dx$$ To bring the limit inside I need to apply the dominated convergence theorem. Keep in mind **I have no knowledge of measure theory...** I was thinking if: $$e^{ix} \le x$$ we can do something. But we cannot compare a complex function with a real function. We **could** set a bound, $x=\pi$ such that: $$ e^{ine^{ix}} \le e^{-in}$$ But we cannot use the dominated convergence theorem, because we a function of $x$, $g(x)$. I cannot prove uniform continuity either. What to do? **The book says we CAN take the limit inside**

**Hint:** $$\left|e^{ine^{ix}}\right|=e^{n\sin x}\leq e^{|n|\sin x}\underset{\text{if }|n|\leq 1}{\leq} e\in L^1(0,\pi)$$

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