hpfan101
  • hpfan101
\[\lim_{x \rightarrow 0^-}(\frac{ 1 }{ x }-\frac{ 1 }{ \left| x \right| })\]
Calculus1
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
When \(x\to0^-\), that means that \(x<0\), so \(|x|=-x\). \[\frac{1}{x}-\frac{1}{|x|}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}\]
hpfan101
  • hpfan101
Oh ok, thank you! So if there's an absolute value and x is less than 0, we just change the sign?
anonymous
  • 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.

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hpfan101
  • hpfan101
Oh I see. Thanks for the explanation!
anonymous
  • anonymous
yw

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