How would you go about proving that:
$\lim_{x\to2^-}\frac{1}{x-2}= -\infty$ (i.e. x approaching 2 from left hand side)
Been struggling to make much headway on this. Not sure how I'm supposed to use Epsilon in this case. So far, I have:
Given any negative number M there exists $\delta$ s.t. $f(x) < M$ if $0<|x-2|<\delta$
-> $\frac{1}{x-2} < M$ if "..."
-> $\frac{1}{M} > x-2$ if "..."
Really not sure how to proceed from here. Can't find any online examples like it and the one in my book is for $\frac{1}{x^2}$ (I understand it, but not sure how to apply it here).
Any help/feedback/advice would be greatly appreciated.