prove $\lim_{x\to 0^-} \frac{x}{|x|} = -1$

Question

I was trying to prove this one, this is how approached it,

As its left hand side limit, For every $\epsilon $ , $\exists \delta$ such that, if $0-\delta < x< 0$ then $|f(x) - L | < \epsilon$.

for given $\epsilon > 0$,

$|\frac{x}{|x|} + 1|<\epsilon$

my thought process somehow i have to convert this expression into "something < x" so that i can choose that "something as $\delta$"

but coming to this problem i thought as limit was approaching from left of 0, so $\frac{x}{|x|}$ will be -1 so substitute into the equation and it's just

|-1 + 1 | and its just 0, which is not the desired result,

how can i move further, any sort of hints or solution .

Answer 1

Say shortly, limit definition is implication in some circumstances. When conclusion is true, then all implication is true. You have $\delta >0 \Rightarrow \varepsilon > 0$, which is tautology, so all your reasonings are correct and exactly they provide the proof.