Verified answer

(==>) By hypothesis, if ε > 0, then there exists N > 0 such that x > N ==> |f(x) - L| < ε.

So, if 0 < x < 1/N, then |f(1/x) - L| < ε.

(<==) If ε > 0, then there exists δ > 0 such that 0 < x < δ ==> |f(1/x) - L| < ε.

Write y = 1x. So, if y > δ, then |f(y) - L| < ε.

I hope this helps!