Prove that $\lim_{x \to 0} \frac{x}{|x|}$ doesn't exist using epsilon delta definition

I'm trying to prove that $\displaystyle\lim_{x \to 0} \dfrac{x}{|x|}$ doesn't exist using epsilon-delta definition. I know i'm supposed to assign a value to epsilon and assume $\displaystyle\lim_{x \to 0} \dfrac{x}{|x|} = L$ and work backwards to get delta. But I'm now stuck on finding delta.

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You don't have to choose a delta. Instead, for each real number – each candidate for the limit – you have to choose an epsilon and show for every positive delta that the delta-neighbourhood of $0$ isn't taken to that epsilon-environment of that candidate. This means, no matter how close you come to $0$, you won't come as close as the chosen $\varepsilon$ to the limit candidate. –  k.stm Jan 20 '13 at 23:09

First note that the function you are considering is: $$f(x) = \frac{x}{\lvert x \rvert} = \begin{cases}1 & \text{for }x>0 \\ -1 & \text{for } x<0\end{cases}.$$ (So the function is not defined at $0$.)

Now let $\epsilon= \frac{1}{2}$ be given. You want to prove that there does not exist an $L\in\mathbb{R}$ and a $\delta > 0$ such that if $\lvert x \rvert < \delta$ then $\lvert f(x) - L\rvert < \epsilon = \frac{1}{2}$. Assume that such did exist. So then for $x\in (-\delta , \delta)$, you would have $$\lvert f(x) - L\rvert < \epsilon = \frac{1}{2}.$$ That is, you have for all $x\in (-\delta, \delta)$ that $$-\frac{1}{2} < f(x) - L < \frac{1}{2}.$$ In particular for $x= \pm\frac{\delta}{2}$ you get the two equations $$-\frac{1}{2} < -1 - L < \frac{1}{2} \\ -\frac{1}{2} < 1 - L < \frac{1}{2}.$$ So you get $$\frac{1}{2} < - L < \frac{3}{2} \\ -\frac{3}{2} < -L < -\frac{1}{2}.$$ But that obviously doesn't work.

Here is my challenge to you: redo this with $\epsilon = 1$.

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Hint: $|x|=-x$ when $x<0$ and $|x|=x$ when $x>0$. Now calculate the limit from the left and from the right. Do they match? What does this imply?

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@Yunhui Shi Another hint: Draw the function $x \mapsto \tfrac{x}{\lvert x \rvert}$, have a look at it, and then choose your $\varepsilon$ wisely. Clayton, he wants to do this by epsilon-delta. –  k.stm Jan 20 '13 at 23:05
@K.Stm.: This argument shows him exactly what $\varepsilon$'s he can pick. From there, this argument validates the $\varepsilon$-$\delta$ proof he wants. –  Clayton Jan 20 '13 at 23:07
Ah, right. That's the hint. –  k.stm Jan 20 '13 at 23:12