[출제의도] 수열의 극한의 성질을 이용하여 극한값을 구할 수 있는가?
주어진 식의 분자와 분모에 n 2 + 3 n + n 2 + n \displaystyle \sqrt{n ^{2} + 3 n} + \sqrt{n ^{2} + n} n 2 + 3 n + n 2 + n lim n → ∞ 1 n 2 + 3 n − n 2 + n \displaystyle \lim\limits _{n \rightarrow \infty} {\frac{1}{\sqrt{n ^{2} + 3 n} - \sqrt{n ^{2} + n}}} n → ∞ lim n 2 + 3 n − n 2 + n 1 = lim n → ∞ n 2 + 3 n + n 2 + n ( n 2 + 3 n ) − ( n 2 + n ) \displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{\sqrt{n ^{2} + 3 n} + \sqrt{n ^{2} + n}}{\left( n ^{2} + 3 n \right) - \left( n ^{2} + n \right)}} = n → ∞ lim ( n 2 + 3 n ) − ( n 2 + n ) n 2 + 3 n + n 2 + n = lim n → ∞ n 2 + 3 n + n 2 + n 2 n \displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{\sqrt{n ^{2} + 3 n} + \sqrt{n ^{2} + n}}{2 n}} = n → ∞ lim 2 n n 2 + 3 n + n 2 + n = lim n → ∞ 1 + 3 n + 1 + 1 n 2 \displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{\sqrt{1 + \frac{3}{n}} + \sqrt{1 + \frac{1}{n}}}{2}} = n → ∞ lim 2 1 + n 3 + 1 + n 1 = 1 + 1 2 \displaystyle = \frac{1 + 1}{2} = 2 1 + 1 = 1 = 1 = 1