I'm looking at $\lim_{x \to \infty} \frac{x}{|x|}$ and my intuition says it's supposed to be 1, which is confirmed by wolframalpha. Yet I'm confused how I'm supposed to prove it's the case when opening up the absolute.
Won't I end up with:
$\lim_{x \to +\infty} \frac{x}{x} = 1$
$\lim_{x \to -\infty} \frac{x}{-x} = -1$
Meaning the limit DNE?
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$\begingroup$As $x\to +\infty$, $x > 0{}{}{}{}{}{}$.
You cannot approach infinity from the right! Infinity is to the right of any, every $x$.
Remember, $x$ is approaching positive infinity, $x$ becomes huge, and we can rest assured that this $x$ increases far beyond the negative reals. So the fact that $x$ can indeed take on negative values, this limit deals with what happens when $x$ increases without bound, not what happens, e.g. as $x\to 0$, in which case the limit would not exist, for the reason you show.
Edit following OP's edit:
$\infty = +\infty$, so you are being asked only to evaluate $$\lim_{x\to +\infty} \frac{x}{|x|}.\tag{1}$$
Note that $\lim_{x\to -\infty} \frac x{|x|}\tag{2}$ is an entirely different limit that the immediately above. The limit of $(1)$ and the limit of $(2)$ both exist. The limit of the first is $1$. The limit of the second is $-1$.
$\endgroup$ $\begingroup$Note that you go either to $\ +\infty$ or to $\ -\infty$, so the value of the limit is +1 in the first case, and -1 in the second.
In this case instead the limit does not exist: $$\ \lim_{x\to0}\frac{1}{x}$$
because it is $+\infty$ for $x\to0^+$, and $-\infty$ for $x\to0^-$
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