Limit of $f(x)=|\log x|$ My textbook solved this problem:
Find $f'(1^{-})$ if
$$f(x)=|\log x|$$
for the interval $x>0$
The textbook solved it by using the method described below:
$$f'(1^{-})=\lim\limits_{x\to 1^{-}} \frac{f(x)-f(1)}{x-1}$$
Which becomes:
$$f'(1^{-})=\lim\limits_{x\to 1^{-}} \frac{|\log x|-|\log 1|}{x-1}$$
They substituted $x=1-h$
$$f'(1^{-})=\lim\limits_{h\to 0^{+}} \frac{|\log (1-h)|-|\log 1|}{-h}$$
Now they claimed the answer to this is $-1$
I really don't understand how they arrived at that answer. Could it be a typo on their part?

An alternative answer which does not require Taylor series or the knowledge of the derivative of $\text{log}$ is to use the limit definition for $\frac{1}{e}$:
$$\frac{1}{e} = \lim_{n\rightarrow \infty} \left( 1- \frac{1}{n} \right)^n = \lim_{m\rightarrow 0} \left( 1- m \right)^{1/m} $$
So that:
$$ -\lim_{h\rightarrow 0} \frac{\left| \text{log}(1-h) \right|}{h}=-\lim_{h\rightarrow 0} \left| \text{log}(1-h)^{1/h} \right| = - \left| \text{log} \left( \lim_{h\rightarrow 0} (1-h)^{(1/h)} \right) \right| $$
$$ = -\left| \text{log} \left(1/e \right) \right| = - \left| -1 \right|=-1 $$
