limit as x approaches infinity+of ((5x^3+7x^2-8x+9))/(11x^2+7x-6)

解

x \to \infty + lim (\frac{( 5 x ^{3} + 7 x ^{2} - 8 x + 9 )}{11 x ^{2} + 7 x - 6})

\infty

解答ステップ

x \to \infty lim (\frac{5 x ^{3} + 7 x ^{2} - 8 x + 9}{11 x ^{2} + 7 x - 6})

最高の分母べき乗で割る:

\frac{5 x + 7 - \frac{8}{x} + \frac{9}{x ^{2}}}{11 + \frac{7}{x} - \frac{6}{x ^{2}}}

= x \to \infty lim (\frac{5 x + 7 - \frac{8}{x} + \frac{9}{x ^{2}}}{11 + \frac{7}{x} - \frac{6}{x ^{2}}})

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} ( 5 x + 7 - \frac{8}{x} + \frac{9}{x ^{2}} )}{lim _{x \to \infty} ( 11 + \frac{7}{x} - \frac{6}{x ^{2}} )}

x \to \infty lim (5 x + 7 - \frac{8}{x} + \frac{9}{x ^{2}}) = \infty

x \to \infty lim (11 + \frac{7}{x} - \frac{6}{x ^{2}}) = 11

= \frac{\infty}{11}

無限大の特性を適用する:

\frac{\infty}{c} = \infty

= \infty