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The limit is

$$ \lim_{x\to0-} (1/x - 1/|x|) $$

I'm teaching myself basic calculus and I don't understand why the limit Does Not Exist. Can someone explain?

August Sepp
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  • You are interested in lim for negative values of $x$, when $x$ is negative, $\frac{1}{|x|}=-\frac{1}{x}$, therefore you limit becomes $\frac{1}{x}+\frac{1}{x}=\frac{2}{x}$ when is $x$ is very very small and negative, $ \frac{2}{x} \to $ very big negative number $"-\infty"$ – Vikram Sep 05 '14 at 04:25
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    Put $x=-1/N$ for $N$ a positive integer. Then $1/x-1/|x|=-2/N$ which gets arbitrarily large and negative. You can either say the limit is $-\infty$ or that it does not exist. – almagest Sep 05 '14 at 04:25

1 Answers1

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The notation $$x \to 0^-$$ indicates that the limit is taken as $x$ approaches $0$ from the left; i.e., $x < 0$. In such a case, recall that for real numbers $x$, the absolute value can be defined as $$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0. \end{cases}$$ Given this, what is $1/|x|$ when $x < 0$? What is $1/x - 1/|x|$ when $x < 0$? Now what is the limit as $x \to 0^-$?

Incidentally, what is the limit as $x \to 0^+$?

heropup
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