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How do I solve [math]\lim_{x\rightarrow 0}\left | \frac{x+1}{x-1} \right |[/math]
That makes sense. And what about when x approaches 0 from the right?
When x approaches 0 from the right, the numerator (x+1) still approaches 1, but now the denominator (x-1) approaches 1. So we get a limit of [math]\lim_{x\rightarrow 0^+}\left | \frac{x+1}{x-1} \right |[/math] = [math]\lim_{x\rightarrow 0^+} \frac{x+1}{x-1}[/math] = -1.