Proving whether a limit exists at a point (piece-wise function) I'm given a function $f(x) = \frac{\lvert x\rvert}{x}$ where $x\neq 0$ and $f(x) = 1$ when $x = 0$. I am asked to prove whether the limit of this function exists as $x\to 0$.
I've tried to use the epsilon-delta method of showing this and it yielded that the limit existed. I am pretty sure my method is incorrect or I am misunderstanding what the actual epsilon-delta method does.
I know you can show the limit exists by taking left and right limits and evaluating the functional value at a point but I would like to know how to do it via the epsilon-delta method.
My proof:
WTS: for all $\varepsilon > 0$ there exists $\delta > 0$ s.t. $\lvert x - 0 \rvert < \delta \implies \lvert \frac{\lvert x\rvert }{x} -1\rvert < \varepsilon.$
As $x$ tends to $0$ positively $\rvert \frac{\lvert x\rvert }{x} -1\rvert$ becomes equal to $0$. As $x$ tends to $0$ negatively $\lvert \frac{\lvert x\rvert }{x} -1\rvert$ becomes equal to $2$.
Hence I am able to find a value for $\delta$ such that for any $\varepsilon > 0$, $\lvert \frac{\lvert x \rvert}{x} -1\rvert$ is less than $\varepsilon$ (as I can let $\delta$ such that $x$ is positive).
Note: now that I have written it in full, is the error that $\lvert x\rvert < \delta$ does not distinguish between positive and negative $x$?
Thanks!
 A: So, I think the misunderstanding you have is that $\varepsilon - \delta$-type proofs cannot be applied to one-sided limits. The first step, when given a function as simple is this, is to sketch it. Just from the sketch itself you should glean the answer to your question: there is a clear discontinuity at $x=0$. We can also see that the function consists of two constant segments. So let's prove that the left- and right-hand limits don't converge by rigorously evaluating both.
To use the $\varepsilon - \delta$ definition we should optimally know the values of the limits beforehand. Luckily, the sketch (or a basic observation) should suggest those pretty quickly, namely:$$\lim_{x\to 0^+} \frac{\lvert x \rvert}{x} = 1$$ and $$\lim_{x\to 0^-}\frac{\lvert x \rvert}{x} = -1.$$ Note that the actual value of $f(x)$ at $0$ isn't even necessary here as we're evaluating limits on a deleted neighbourhood (unless we want to additionally show the function is discontinuous). So, let's do both proofs from scratch by first reciting the definitions. For the first one we have that the right-hand limit is $1$ if and only if:$$\forall \varepsilon > 0, \exists \delta > 0 \text{  s.t. } 0<x<0+\delta \implies \lvert \frac{\lvert x \rvert }{x} - 1\rvert < \varepsilon.$$ In this case we don't really need to choose our $\delta$ in terms of $\varepsilon$ or otherwise as we're dealing with a constant function (for all $x>0$), hence $$\lvert \frac{\lvert x \rvert }{x} - 1\rvert = \lvert \frac{x - x}{x} \rvert = 0 < \varepsilon .$$ Note that a similar condition will hold for any constant function. We're also not worried about dividing by $0$ because of the restriction $0<x$ in the antecedent.
The proof for the left-hand limit is done analagously. We have to show:$$\forall \varepsilon >0, \exists \delta > 0, \text{ s.t. } 0-\delta < x < 0 \implies \lvert \frac{\lvert x \rvert}{x} - (-1)\rvert < \varepsilon.$$ Again, we don't have to bother about the value of $\delta$ here as $$ \lvert \frac{\lvert x\rvert}{x} + 1 \rvert = \lvert  \frac{\lvert x \rvert + x}{x} \rvert = 0 < \varepsilon$$ because $\lvert x \rvert + x = 0$ due to having the restriction $x < 0$ in our antecedent.
This way we have rigorously proved that the right- and left-side limits don't converge to a single value, therefore the limit does not exist.
