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**novice** Prove that $\displaystyle \lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Proof:

Let $\displaystyle f(x)=\frac{|x|}{x}. $ Assume, to the contrary, that $\displaystyle \lim_{x \to 0} f(x)$ exists. Then there exists a real number $\displaystyle L$ such that $\displaystyle \lim_{x \to 0} f(x) = L$. Let $\displaystyle \epsilon = 1$. Then there exists $\displaystyle \delta >0$ such that if $\displaystyle x$ is a real number satisfying $\displaystyle 0<|x-0|<\delta$, then $\displaystyle |f(x)-L|<\epsilon$.

***** If a limit L exists, then f(x) approaches that limit, hence f(x)-L must be at least <1 at some point*******

We consider two cases.

Case 1. $\displaystyle L\geq0$. Consider $\displaystyle x=- \delta/2$.

******But if $\displaystyle L\ge0,$ x must be $\displaystyle \ge0$******

Then $\displaystyle |x|=\delta/2 <\delta$. However, $\displaystyle f(x)=f(-\delta/2) =(\delta/2)/(-\delta/2)=-1$. So $\displaystyle |f(x)-L|=|-1-L|=1+L > 1$, a contradiction.

******$\displaystyle f(x)=1, |f(x)-L|=|1-L|$********

Case 2. $\displaystyle L<0$. Consider $\displaystyle x=\delta/2$.

******$\displaystyle x=-\frac{\delta}{2}$*******

Then $\displaystyle |x|=\delta/2 <\delta$. Also, $\displaystyle f(x)=f(\delta/2) =(\delta/2)/(\delta/2)=1$.

******$\displaystyle f(x)=-1$*******

So $\displaystyle |f(x)-L|=|1-L|=1-L > 1$, a contradiction.

******$\displaystyle |f(x)-L|=|-1-L|******$

Remarks: I am particularly troubled by the choice of $\displaystyle x$'s being contrary to the values of $\displaystyle L$. Intuitively, we know that the value of $\displaystyle x$ cannot be negative when $\displaystyle L+=\lim_{x \to 0^+}f(x)$ and $\displaystyle x$ cannot be positive when $\displaystyle L-=\lim_{x \to 0^-}f(x)$.

Please tell me where I am wrong.