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

解答

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

\infty

求解步骤

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

除以分母的最高次幂:

\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

不定型除外

= \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}

运用无穷大属性:

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

= \infty

\int sin^{4} (7 x) d x

in v erse l a pl a ce \frac{1}{( s - 5 ) ^{2}}

\frac{d}{d x} (x 1 + x^{2})

y^{^{''}} + 4 y^{^{'}} + 13 y = 0

a re a f (x) = 2 x - 4, 3, 6