Question

\[\lim_{x \rightarrow 0^-}(\frac{ 1 }{ x }-\frac{ 1 }{ \left| x \right| })\]

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

Oh I see. Thanks for the explanation!

Answer 5

yw