Why is $|x| = -x$ when $x \rightarrow -\infty $ I really despise maths, so bear with me. I'm not the greatest at it.
I emailed my Maths1003 (Calc I) prof to ask a question, and at least to me, he answered it in a riddle.
Question is: Evaluate the Limit: $$\lim\limits_{x \to -\infty} \frac{|x|+x}{x+1}$$
But I don't understand how the absolute value can be negative. Doesn't that defeat the point of it being absolute?
Answer is $0$.
He did say that $|x| = \sqrt{x^2}$, which I do understand.
Help!
 A: 
But I don't understand how the absolute value can be negative.

It's not. If $x$ is negative then $-x$ is positive.
For example:
$$
|{-2}| = -(-2) = 2.
$$
You can't conclude that a quantity is negative just because you see a
 "$-$" sign in front of it.
A: Hint. By definition of absolute value, if $x<0$ then $|x|=-x$. Therefore, for $x<0$, the sum $|x|+x$ is identically zero (and your limit is zero too). 
Note that $-x$ is not always a negative number. It is negative if and only if $x$ is positive. 
A: Since of interest is the limit of the function as $x \to -\infty$; it suffices to  check for $x < 0$. For, by definition, that we want to know to what number the function "stays around" for $x$ goes below every bound $<0$. Considering $x < 0$, we immediately notice that $|x| + x = 0$. Since the function is always $=0$ for $x < 0$, its limit as $x \to -\infty$ is $0$.
It is because it suffices to consider $x < 0$ in this case that you see $|x| = -x$. The reason is that if $x < 0$ then $-x > 0$; and $|x| = -x$ for all $x < 0$.
