limn→∞∣[2(n+1)]![(n+1)!]2x2(n+1)(2n)!(n!)2x2n∣=4∣x∣2\lim_{n \to \infty} |\frac{\frac{[2(n+1)]!}{[(n+1)!]^2}x^{2(n+1)}}{\frac{(2n)!}{(n!)^2}x^{2n}}|=4|x|^2limn→∞∣(n!)2(2n)!x2n[(n+1)!]2[2(n+1)]!x2(n+1)∣=4∣x∣2
这玩意怎么把两边绝对值调到和分数一样
