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:
$x &lt; 0$
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 $2(1-x) &lt; ε$
$\implies 1-x &lt; ε/2$
$\implies x&gt;1-ε/2$
I have to prove that $ε/2&lt;x&lt;0$  

Where did I go wrong?
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 $|x|&lt;\epsilon/2$, and $x$ is negative, then $|y-(-1)|&lt;\epsilon$. (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 $\lim_{x\to 0^-} (2x-1)=-1$.
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:

$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}=\lim_{x \to 0^-} \frac{o+\frac{-x}{x}}{1}=\lim_{x \to 0^-}\frac{0-1}{1}=-1$$