Can the order of absolute value and limit be swapped?

Question

Is $|\lim_{n\to\infty}f_n(x)|$ equal to $\lim_{n\to\infty}|f_n(x)|$, given that $\lim_{n\to\infty} f_n(x)$ exists? ($f_n:\mathbb{R} \rightarrow \mathbb{C}$).

Answer 1

Your question goes to the proof of the following:

Result: If $\lim_{n\to\infty}a_n=L$ then $\lim_{n\to\infty}|a_n|=|L|$.

To prove this, let $\epsilon>0$. Then we can find $N\in\Bbb N$ such that for all $n\geq N$ then $$|a_n-L|<\epsilon.$$ But we know that for such $n\geq N$ (in fact for all $n$) $$\bigg||a_n|-|L|\bigg|\leq |a_n-L|.$$ Thus, for all $n\geq N$, we get $$\bigg||a_n|-|L|\bigg|<\epsilon.$$ This proves the statement.

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In your question, let us write

$$\lim_{n\to\infty}f_n(x)=L(x).$$

Using the result above, we get $$\lim_{n\to\infty}|f_n(x)|=|L(x)|=|\lim_{n\to\infty}f_n(x)|$$