Given the equation: $$(x-1)\log|x-1|$$ Determine the left and right limits about $x = 1$: $$\lim\limits_{x\to 1} (x-1)\log|x-1|= \lim\limits_{x\to 1} \frac{\log|x-1|}{1/(x-1)}$$ left limit: $$\lim\limits_{h\to 0} \frac{\log|(1-h)-1|}{1/((1-h)-1)} = \lim\limits_{h\to 0} \frac {\log(h)}{-1/h} = \lim\limits_{h\to 0} \frac {1/h}{1/h^2} = 0$$
There is a similar expression for the right limit with $(1-h)$ replaced with $(1+h)$
My questions are:
How do you determine $x = 1 - h$ for the left (right) limits respectively?
Is it simply looking at where the denominator would equal 0? Or does it have to do with the limit $x \to 1$ such that when $h \to 0$, $x = 1$?
For example, what would the expression for $x$ be if $\lim\limits_{x\to 5}$ instead? Is this even relevant?
Why is $x = 1 - h$ for the left and $x = 1 + h$ for the right?
First question: $x = 1-h$ from the left because the limit approaching from the left only involves values of $x$ less than $1.$ Similarly, the limit from the right only involves values of $x$ greater than $1.$
Second question: It somewhat relates to when the denominator is zero. This is because direct substitution cannot be used when $x\to 1.$
Third question: It relates to the fact that as $h\to 0^+$ (which is implied in your question), $1-h \to 1^-$ and $1+h\to 1^+,$ and hence the values $1-h$ and $1+h$ for the left and right limits respectively.
Fourth question: If the limit is evaluated at $5,$ then this method would be much less effective as direct substitution could be used.
Fifth question: If the limit is evaluated at $5,$ then this would be a completely different question.
Sixth question: I've already answered this.
Hope this clears things up.;)