limit as x approaches infinity of (7x-1)/(\sqrt[3]{5x^3+4x-1)}

解

x \to \infty lim (\frac{7 x - 1}{3 5 x ^{3} + 4 x - 1})

\frac{7}{3 5}

十進法表記

4.09362 \dots

解答ステップ

x \to \infty lim (\frac{7 x - 1}{3 5 x ^{3} + 4 x - 1})

最高の分母べき乗で割る:

\frac{7 - \frac{1}{x}}{3 5 + \frac{4}{x ^{2}} - \frac{1}{x ^{3}}}

= x \to \infty lim \frac{7 - \frac{1}{x}}{3 5 + \frac{4}{x ^{2}} - \frac{1}{x ^{3}}}

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 _{x \to \infty} ( 7 - \frac{1}{x} )}{lim _{x \to \infty} ( 3 5 + \frac{4}{x ^{2}} - \frac{1}{x ^{3}} )}

x \to \infty lim (7 - \frac{1}{x}) = 7

x \to \infty lim (3 5 + \frac{4}{x ^{2}} - \frac{1}{x ^{3}}) = 35

= \frac{7}{3 5}