show 1/x is continuous at x = c > 0
work:
given e>0, consider
|1/x - 1/c| < e
1/c - e < 1/x < 1/c+e
(1-ce)/c < 1/x < (1+ce)/c
c/(1+ce) < x < c/(1-ce)
c/(1+ce)-c < x - c < c/(1-ce) - c
(c-c-(c^2)e)/(1+ce) < x - c < (c-c+(c^2)e)/(1-ce)
-(c^2)e/(1+ce) < x - c < (c^2)e/(1-ce)
But how to get this in a |x-c|< g(e), where g(e) is some function dependent on e?
I have seen other approaches for this proof, but I am a bit annoyed I cannot make the above approach work.
Thanks,