hpfan101 one year ago $\lim_{x \rightarrow 0^-}(\frac{ 1 }{ x }-\frac{ 1 }{ \left| x \right| })$

1. anonymous

When $$x\to0^-$$, that means that $$x<0$$, so $$|x|=-x$$. $\frac{1}{x}-\frac{1}{|x|}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}$

2. hpfan101

Oh ok, thank you! So if there's an absolute value and x is less than 0, we just change the sign?

3. anonymous

Kind of, it depends on the values of $$x$$ you're considering. The absolute value of a number is defined by $|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}$ This means whenever $$x$$ is some positive number, its absolute value is the same number, i.e. $$|x|=x$$. If $$x$$ is a negative number, then the absolute value will cancel that sign to make it positive. But it wouldn't be true that $$|x|=x$$ because $$x$$ is negative. For example, the previous equation would suggest that $$|-2|=-2$$, but that's not true. We make up for the sign by making $$|x|=-x$$ whenever $$x$$ negative.

4. hpfan101

Oh I see. Thanks for the explanation!

5. anonymous

yw