limit as n approaches infinity of (1+\sqrt[3]{n})/(1+\sqrt[3]{n+1)}

解

n \to \infty lim (\frac{1 + 3 n}{1 + 3 n + 1})

1

解答ステップ

n \to \infty lim (\frac{1 + 3 n}{1 + 3 n + 1})

3 n + 1

で割る

: \frac{\frac{1}{3 n + 1} + 3 \frac{n}{n + 1}}{\frac{1}{3 n + 1} + 1}

= n \to \infty lim (\frac{\frac{1}{3 n + 1} + 3 \frac{n}{n + 1}}{\frac{1}{3 n + 1} + 1})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

不定形式の場合を除く

= \frac{lim _{n \to \infty} ( \frac{1}{3 n + 1} + 3 \frac{n}{n + 1} )}{lim _{n \to \infty} ( \frac{1}{3 n + 1} + 1 )}

n \to \infty lim (\frac{1}{3 n + 1} + 3 \frac{n}{n + 1}) = 1

n \to \infty lim (\frac{1}{3 n + 1} + 1) = 1

= \frac{1}{1}

簡素化

= 1