推 pobm:謝謝^^ 140.122.61.215 07/27 15:04
※ 引述《GayerDior (蠟筆小新<( ̄. ̄)/)》之銘言:
: E:
: 証明下列二敘述是對等的。
: ^^^^
: (1) lim f(x)=L ; (2) lim f(a+h)=L 。
: x->a f->0
: 如果有人會寫,
: 請幫我寫出詳細過程,
: 謝謝唷~~~ \(╯▽╰)/
設 {g(y) : y in D(g)} 包含在 D(f), 若
1) f(x) -> L as x -> a
2) g(y) -> a as y -> b
則 f(g(y)) -> L as y -> b
for any e>0, there exists d>0 such that
|f(x) - L| < e whenever 0 < |x-a| < d
for this d, there exists d'>0 such that
|g(y) - a| < d whenever 0 < |y-b| < d'
then |f(g(y)) - L| < e whenever 0 < |g(y)-a| < d
thus |f(g(y)) - L| < e whenever 0 < |y-b| < d'
so f(g(y)) -> L as y -> b
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朋友,風起了,蟬鳴了,你聽見了嗎。
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