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I know my answer is correct, but I am unsure if my steps are correct.

I'm asked to find the limit $\lim_{x \to 0} = \displaystyle\frac{2x - |x|}{|3x| - 2x}$

I start by jotting down some basics about $|x|$

\begin{equation*} |x| = \begin{cases} x, & \text{if $x\ge0$}.\\ -x, & \text{if $x < 0$}. \end{cases} \end{equation*}

And then proceed to solve the limit

\begin{equation*} \lim_{x \to 0} = \displaystyle\frac{2x - |x|}{|3x| - 2x}\\ \lim_{x \to 0} = \displaystyle\frac{2x - x}{3x - 2x}\\ = \displaystyle\frac{2 - 1}{3 - 2}\\ = 1 \end{equation*}

As pointed out, I now look at the limit from the negative side and get \begin{equation*} \lim_{x \to 0} = \displaystyle\frac{2x - |x|}{|3x| - 2x}\\ \lim_{x \to 0} = \displaystyle\frac{2x - (-x)}{-3x - 2x}\\ = \displaystyle\frac{2 + 1}{-3 - 2}\\ = -\displaystyle\frac{3}{5} \end{equation*}

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    $\begingroup$ You haven't considered $x < 0$ yet. $\endgroup$ – Daniel Fischer Mar 4 '14 at 20:06
  • $\begingroup$ Looking at this I think I should show that I divide by $x$ $\endgroup$ – Leon Mar 4 '14 at 20:07
  • $\begingroup$ That much is clear. But do look at $x < 0$. $\endgroup$ – Daniel Fischer Mar 4 '14 at 20:08
  • $\begingroup$ Do I have to consider $x < 0$ when the limit approaches 0? I guess I have to else you wouldn't point that out $\endgroup$ – Leon Mar 4 '14 at 20:08
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    $\begingroup$ Since the domain contains positive as well as negative sequences converging to $0$, you need to look at the approach to $0$ from both sides. $\endgroup$ – Daniel Fischer Mar 4 '14 at 20:09
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If $x> 0$, then $\frac{2x-|x|}{|3x|-2x}=\frac{2x-x}{3x-2x}=\frac{x}{x}=1$. If $x<0$ then $\frac{2x-|x|}{|3x|-2x}=\frac{2x-(-x)}{-3x-2x}=-\frac{3}{5}$. By using these identities if follows that the right hand limit and left hand limit of the quotient does not agree, and hence the limit does not exist. Note that the quotient is not defined at $x=0$. (Look up the definition of the limit of a function)

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