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

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

Calculus1
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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When \(x\to0^-\), that means that \(x<0\), so \(|x|=-x\). \[\frac{1}{x}-\frac{1}{|x|}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}\]
Oh ok, thank you! So if there's an absolute value and x is less than 0, we just change the sign?
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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Oh I see. Thanks for the explanation!
yw

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