limit as x approaches infinity of (\sqrt[3]{8x^9+x^6+2})/(4x^3+x^2-1)

解

x \to \infty lim (\frac{3 8 x ^{9} + x ^{6} + 2}{4 x ^{3} + x ^{2} - 1})

\frac{1}{2}

十進法表記

0.5

解答ステップ

x \to \infty lim (\frac{3 8 x ^{9} + x ^{6} + 2}{4 x ^{3} + x ^{2} - 1})

最高の分母べき乗で割る:

\frac{3 8 + \frac{1}{x ^{3}} + \frac{2}{x ^{9}}}{4 + \frac{1}{x} - \frac{1}{x ^{3}}}

= x \to \infty lim \frac{3 8 + \frac{1}{x ^{3}} + \frac{2}{x ^{9}}}{4 + \frac{1}{x} - \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} ( 3 8 + \frac{1}{x ^{3}} + \frac{2}{x ^{9}} )}{lim _{x \to \infty} ( 4 + \frac{1}{x} - \frac{1}{x ^{3}} )}

x \to \infty lim (3 8 + \frac{1}{x ^{3}} + \frac{2}{x ^{9}}) = 2

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}