AL Pure Maths

I think the argument of EMK is not good enough
We need to think thatthe range of x is bounded [0,1/√2] . So, p(x) is bounded from above and below
i.e. C<=p(x)<=M
Thus C∫(0 to 1/√2) x^k dx <= ∫(0 to 1/√2) [x^kp(x)]dx <= M∫(0 to 1/√2) x^k dx

Therefore to prove your statement, it suffices to show that
lim(k→∞) ∫(0 to 1/√2) x^k dx = 0

Now since
∫(0 to 1/√2) x^k dx = 1/[(k+1)(√2)^{k+1}]
and it is clear that
lim(k→∞) 1/[(k+1)(√2)^{k+1}] = 0

Hence we have
lim(k→∞) ∫(0 to 1/√2) x^k dx = 0
and thus the original statement holds.

某程度上來說，myisland8132 的做法是比較「正路」，不過EMK 的方法也是可以，所以我不會說是"not good enough".
(無意冒犯myisland8132，純綷討論數學。)
至於Dominated Convergence Theorem，實在太強了，所謂殺雞焉用牛刀，個人認為在做elementary的問題時，能避免用太強的定理是好的，有助掌握基本功夫。

Dominated Convergence Theorem
http://en.wikipedia.org/wiki/Dominated_convergence_theorem

2009-02-13 21:31:29 補充：
: )

Since the integral is linear, that means, for two polynomials p(x) and q(x),
∫(0 to 1/√2) [p(x)+q(x)]dx = ∫(0 to 1/√2) p(x)dx + ∫(0 to 1/√2) q(x)dx

Therefore to prove your statement, it suffices to show that
lim(k→∞) ∫(0 to 1/√2) x^k dx = 0

Now since
∫(0 to 1/√2) x^k dx = 1/[(k+1)(√2)^{k+1}]
and it is clear that
lim(k→∞) 1/[(k+1)(√2)^{k+1}] = 0

Hence we have
lim(k→∞) ∫(0 to 1/√2) x^k dx = 0
and thus the original statement holds.