Limit of a difference

Question

Let $\lim_{n \to \infty} f_n(x) = f(x)$. Now consider $$\lim_{n \to \infty} (f_n(x) - f(x))$$ Usually I would say that $$\lim_{n \to \infty} (f_n(x) - f(x)) = \lim_{n \to \infty} f_n(x) - \lim_{n \to \infty} f(x) = 0$$ but what happens if $f(x) \to \infty$ for $x \to \infty$? Is it still possible to decompose the limit in the above fashion?

Answer 1

You can't have $f(x)\to \infty$ for some $x$. $f(x)$ is a fixed value and not something that can "tend" anywhere. When you say $f(x)=\lim_{n\to\infty} f_n(x)$ you say (I hope, at least implicitly) that the limit $\lim_{n\to\infty} f_n(x)$ exists (for all $x$). However, if for some $xn \Bbb R$, the sequence with $n$th term $f_n(x)$ is divergent, then $f(x)$ is not defined in the first place.

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However, if you work in the extended reals and it may happen for some $x$ that $f_n(x)$ diverges to $\infty$, then you may assign $f(x)=\infty$. (Note that this still won't help for cases where $f_n(x)$ oscillates, for example). In that case we have (with all $f_n(x)$ being finite real numbers)

$$ \lim_{n\to\infty}(f_n(x)-f(x))=\lim_{n\to\infty}(f_n(x)-\infty)=\lim_{n\to\infty}(-\infty)=-\infty$$

How can our trusted rules for addition of limits fail? Well, they don't because the sequence $f_n(x)$ does not converge! It is always dangerous and strictly speaking at least an abuse of notation if one writes $\lim_{n\to\infty}f_n(x)=\infty$ for the case of a divergence to $\infty$.