Find left hand limit of a function $f(x)=\frac{2x^2+|x|}{x}$, 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 $ε/2<x<0$  
Where did I go wrong?
 A: There is a little minus sign glitch in the calculation.  In the displayed line, we should have 
$$|y-(-1)|=|y+1|=|(2x-1)+1|=|2x|.$$
Now it is easy to see that if $|x|<\epsilon/2$, and $x$ is negative, then $|y-(-1)|<\epsilon$. (We need to have $x$ negative so that the expression $2x-1$ for $y$ will be correct.)
Comment: I assume that you are expected to use "$\epsilon$-$\delta$." But at the informal level, it is clear that $\lim_{x\to 0^-} (2x-1)=-1$. 
A: When $x<0$ then $|x|=-x$. Therefore one has for all $x<0$ the equality
$$f(x)={2x^2-x\over x}=2x-1\ ,$$
and as the right side is continuous at $x=0$ it follows that
$$\lim_{x\to0-}f(x)=(2x-1)\bigr|_{x=0}=-1\ .$$
A: My solution:
First of all devide numerator and denominator by $x$
$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}$$
Observe that we are interested only in case when $|x|=-x$ so we may write:
$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}=\lim_{x \to 0^-} \frac{o+\frac{-x}{x}}{1}=\lim_{x \to 0^-}\frac{0-1}{1}=-1$$
