# Absolute value of limit can be taken outside

calculuslimitsreal-analysis

We have a sequence of functions $$f_n$$ that converge to $$f$$.

As part of a proof I am writing, if I can prove that

$$|f(x) – f(y)| = \lim_{n \to \infty}|f_n(x) – f_n(y)|$$

then I am done. Proving this requires me to show that

$$|\lim (f_n(x) – f_n(y)) | = \lim |f_n(x) – f_n(y)|$$

and I'm not sure how I can do that. I found this answer limit of absolute value of something equals absolute value of limit of something? which says it exists by 'continuity of absolute value' although I am not sure how that justifies it. Could someone elaborate?

We note that if $$x_n \to x$$ and $$g$$ is continuous, then: $$\lim_{n \to \infty} g(x_n) = g\left(\lim_{x \to \infty} x_n\right)$$ i.e. the limit and the function can be interchanged. The absolute value function $$x \mapsto |x|$$ is continuous. Therefore, in your case we have: \begin{align*} |f(x) - f(y)| &= \left|\left(\lim_{n \to \infty} f_n(x)\right) - \left(\lim_{n \to \infty} f_n(y)\right)\right| \\ &= \left|\lim_{n \to \infty} (f_n(x) - f_n(y))\right| \\ &= \lim_{n \to \infty} |f_n(x) - f_n(y)| \end{align*}