limit as x approaches infinity of (sqrt(7x^3+8x))/(4x-3)

解答

x \to \infty lim (\frac{7 x ^{3} + 8 x}{4 x - 3})

\infty

求解步骤

x \to \infty lim (\frac{7 x ^{3} + 8 x}{4 x - 3})

除以分母的最高次幂:

\frac{x ^{\frac{1}{2}} \frac{8}{x ^{2}} + 7}{4 - \frac{3}{x}}

= x \to \infty lim \frac{x ^{\frac{1}{2}} \frac{8}{x ^{2}} + 7}{4 - \frac{3}{x}}

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} ( x ^{\frac{1}{2}} \frac{8}{x ^{2}} + 7 )}{lim _{x \to \infty} ( 4 - \frac{3}{x} )}

x \to \infty lim (x^{\frac{1}{2}} \frac{8}{x ^{2}} + 7) = \infty

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

= \frac{\infty}{4}

运用无穷大属性:

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

= \infty

\frac{d}{d x} (2 x - x^{3})

\frac{d}{d x} (x^{2} + y^{2} + z^{2} + x y)

x y^{2} + 3 y^{2} - x^{2} y^{^{'}} = 0

\frac{d}{d x} (\frac{3 x - 2}{2 x - 3})

\int \frac{1}{t ^{\frac{3}{2}}} d t