limit as x approaches infinity of (9x^3+x^2-5)/(3x^4+4x^2+3)

解

x \to \infty lim (\frac{9 x ^{3} + x ^{2} - 5}{3 x ^{4} + 4 x ^{2} + 3})

0

解答ステップ

x \to \infty lim (\frac{9 x ^{3} + x ^{2} - 5}{3 x ^{4} + 4 x ^{2} + 3})

最高の分母べき乗で割る:

\frac{\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}}{3 + \frac{4}{x ^{2}} + \frac{3}{x ^{4}}}

= x \to \infty lim (\frac{\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}}{3 + \frac{4}{x ^{2}} + \frac{3}{x ^{4}}})

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} ( \frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}} )}{lim _{x \to \infty} ( 3 + \frac{4}{x ^{2}} + \frac{3}{x ^{4}} )}

x \to \infty lim (\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}) = 0

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

= \frac{0}{3}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0