$f(x)=\frac{2x^2+|x|}{x}$, show that $\lim_{x\to 0^-} f(x)=-1$
My solution:
$x < 0$
Step 1: Therefore $y=(2x^2-x)/x$
Step 2: $y=2x-1$
Step 3: As $x$ approaches $0$ from left, the difference between $0$ and $x$ diminishes $\implies$ $y$ approaches $-1$ from left and the difference between $-1$ and $y$ diminishes
Therefore $|y-(-1)|=1-y=1-(2x-1)=2-2x=2(1-x)$ Step 4: Let $2(1-x) < ε$
$\implies 1-x < ε/2$
$\implies x>1-ε/2$
I have to prove that $ε/2
Where did I go wrong?