Alright, I feel like this is simpler than I think it is...
So the limit we want to know is $$\lim_{x \to 0} f(x)=|x|$$
First we look at the piecewise definition which is given by $$|x| = \begin{cases} x, & x\geqslant 0 \\ -x, & x\lt0 \end{cases}$$
We then look at the one sided limits, for the limit to 0 from above, we consider the case where $$x\geqslant 0$$ such that $$\lim_{x \to 0^+} x$$ Yet this leaves us with just an x, which as it goes to 0... is 0? Yet the solutions I have calculate it in the followin way, $$\lim_{x \to 0^+} \frac{|x|}{x} = 1$$
Why is it divided by x, where does that come from? And furthermore, why does it approach 1 when there are other numbers that exist between 1 and 0 that x could approach... I guess I'm also missing an intuitive understanding?
In order for the limit to exist, you need the side limits to commute. In this case :
$$\lim_{x \to 0} |x| = \lim_{x \to 0} |x| = \begin{cases} \lim_{x\to 0^+} x, \space x \geq 0 \\ \lim_{x \to 0^-} -x, \space x<0\end{cases} = \begin{cases} 0, \space x \geq 0 \\ 0, \space x <0 \end{cases}$$
which means that the side limits are equal and thus : $\lim_{x \to 0} |x| = 0$.
As far as the $\frac{|x|}{x}$ part, if you divide something by a quantity, you will also need to multiply it by that. This is why you're getting $1$ as a wrong answer there.