I'm reading this proof on Rudin Real and Complex analysis exercise 3.5 (d), and got fairly confused here:
$f$ is a complex measurable function defined on $X$, and in this problem, we required that $\|f\|_r < \infty$ for some $r > 0$. Given a sequence $\{p_n\}$ such that $p_n \to 0$, why is the sequence $\{f_n\}$ defined by $f_n := \frac{|f(x)|^{p_n} - 1}{p_n} - \log f(x)$ convergent to $0$?
Notice $$\lim_{n\rightarrow\infty} \frac{|f(x)|^{p_n}-1}{p_n} = \frac{|f(x)|^{0}-1}{0} = \frac{1-1}{0} = \frac{0}{0}$$
Let's use l'Hôpital rule for $p_n$:
$$ \frac{|f(x)|^{p_n} \log(|f(x)|)}{1}$$
Then,
$$ \lim_{n\rightarrow\infty} \frac{|f(x)|^{p_n}-1}{p_n} = \lim_{n\rightarrow\infty}\frac{|f(x)|^{p_n} \log(|f(x)|)}{1} =\frac{|f(x)|^{0} \log(|f(x)|)}{1}=\frac{ \log(|f(x)|)}{1} $$
Concluding, $$\lim_{n\rightarrow\infty} f_n:=\lim_{n\rightarrow\infty}(\frac{|f(x)|^{p_n}-1}{p_n}−\log f(x))=\lim_{n\rightarrow\infty}(\frac{|f(x)|^{p_n}-1}{p_n})−\lim_{n\rightarrow\infty}\log f(x))=\frac{ \log(|f(x)|)}{1} - \log(|f(x)|) =0$$
