20/06/24 09:51:26.93 NoItLLRp.net
>>593
f(x) := 1/(x^2+1)
|f(x+h)-f(x)| = ... = |(2hx + hh)/{((x+h)^2+1)(x^2+1)}|
≦ |2hx|/|((x+h)^2+1)(x^2+1)| + |hh|/|((x+h)^2+1)(x^2+1)|
≦ |2hx|/|x^2+1| + |hh| {∵ |1/((x+h)^2+1)|≦1, 1/|((x+h)^2+1)(x^2+1)|≦1}
≦ |h| + |hh| {∵ |2x/x^2+1|≦1}
≦ (|h|+1/2)^2 - 1/4
∀ε>0, ∃δ = √(ε+1/4) - 1/2
|h| < δ ⇒ |f(x+h)-f(x)| < ε