$\lim_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right),$ the limit does not exist. Is this a good justification?

$$\lim_{x \to 0^-} \left(\frac{1}{x} -  \frac{1}{|x|}\right),$$ the limit does not exist.

Because
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$ 
And
$$\lim_{x \to 0^-} \frac{1}{|x|} = \infty.$$
We know that
$$\lim_{x \to 0^-} \left(\frac{1}{x} -  \frac{1}{|x|}\right) = -\infty.$$ Hence the limit does not exist.
 A: No, this is not a good justification. By similar logic, $4=\lim_{x\to\infty} 4= \lim_{x\to\infty} (x+4)-x=\lim_{x\to\infty} x+4 +\lim_{x\to\infty} -x=\infty-\infty$. It is only valid to break up a limit into a sum when the values are finite.
Hint: what is $|x|$ when $x<0$? Simplify the expression inside the parenthesis using this.
A: Your justification is not sufficient, since limit laws generally only apply when the limits are finite. 
The correct way to handle the limit is to note that since $x < 0$, $|x| = -x$. Hence,
$$\frac{1}{x} - \frac{1}{|x|} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x} \to -\infty$$
as $x \to 0^-$.
A: You can only use that the limit of the difference is the difference of the limits when the limit is finite. To argue this note that we are approaching $0$ from the left. So all values of $x$ will be negative. Hence $|x|=-x$. So
$\lim_{x\to0^{-}}\bigg(\frac{1}{x}-\frac{1}{|x|}\bigg)=\lim_{x\to0^{-}}\frac{2}{x}=-\infty$
A: For a limit to exist the lateral limits have to be the same.
