極限 $\lim_{n \to \infty} \frac{1+n}{1+n^2}$ を求める問題です。解析学極限数列の極限2025/4/221. 問題の内容極限 limn→∞1+n1+n2\lim_{n \to \infty} \frac{1+n}{1+n^2}limn→∞1+n21+n を求める問題です。2. 解き方の手順分子と分母を n2n^2n2 で割ります。limn→∞1+n1+n2=limn→∞1n2+nn21n2+n2n2\lim_{n \to \infty} \frac{1+n}{1+n^2} = \lim_{n \to \infty} \frac{\frac{1}{n^2} + \frac{n}{n^2}}{\frac{1}{n^2} + \frac{n^2}{n^2}}limn→∞1+n21+n=limn→∞n21+n2n2n21+n2n=limn→∞1n2+1n1n2+1= \lim_{n \to \infty} \frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n^2} + 1}=limn→∞n21+1n21+n1n→∞n \to \inftyn→∞ のとき 1n→0\frac{1}{n} \to 0n1→0 かつ 1n2→0\frac{1}{n^2} \to 0n21→0 なので、limn→∞1n2+1n1n2+1=0+00+1=01=0\lim_{n \to \infty} \frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n^2} + 1} = \frac{0+0}{0+1} = \frac{0}{1} = 0limn→∞n21+1n21+n1=0+10+0=10=03. 最終的な答え0