If then .
So let .
Hi, I understand thus but I'm struggling with the proof.
I have so far:
f(5) = 1/5
so we need to show that for all positive epsilon(E) such that that there exists positive delta(d) s.t |(1/x) - 1/5)| < E for all x satisfying |x-5|<d
Then when I searched some threads on here I saw something that I think might help, but this is what it seemed to imply anyway :
Now |(1/x) - (1/5)| = |x-5| / |x||5|
However d>0 so when |x-5| = (1/2).|5| by the triangle law we would have
|x|=> 5-|x-5| >(1/2).|5|
therefore d <= (1/2) . |5|
I don't really see where the bit in red came from, which is why I'm not sure what I have found is correct, and I am also unsure how to finish off the proof. Any help would be great thank you