四川大学历年硕士研究生入学考试数学分析试题解析之极限篇.pdfVIP

1Æ No.1 2013 10 Oct, 2013  gaodengshuxue16888@163.com    1 1 lim (1 + x)x = e x→0 x 2 1 2.( 2011 ) lim (e + x + 3 sin x)2x x→0 x 2 1 e +x +3 sin x−1 = lim [1 + (ex + x2 + 3 sin x − 1)]ex+x2+3 sin x−1 2x x→0 x 2 x e +x +3 sin x−1 e +2x+3cos x 2 = lim e 2x = lim e 2 = e x→0 x→0 ln(ex + x2 + 3 sin x) ex + 2x + 3 cos x lim = lim = 2 x→0 2x x→0 2(ex + x2 + 3 sin x) x 2 1 2 lim (e + x + 3 sin x) 2x = e x→0 2 Æ {xn }, {yn } {zn } (1) ∃n0 ∈ N, n n0 yn ≤ xn ≤ zn , (2) lim yn = a, lim zn = a. n→∞ n→∞ {xn }  lim xn = a. n→∞ n kπ π cos 3.( 2000 ) lim sin n kπ n→∞ n 2 + sin k=1 n 2 1Æ n n n