limit as h approaches infinity of (3h+2xh^2+x^2+h^3)/(4-3xh-2x^3h^3)

解答

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

- \frac{1}{2 x ^{3}}

求解步骤

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

除以分母的最高次幂:

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

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

不定型除外

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

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

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

= \frac{1}{- 2 x ^{3}}

化简

= - \frac{1}{2 x ^{3}}

x \to 3 lim (sin (\frac{1}{x - 3}))

\frac{d}{d x} ((x^{2} + 5)^{\frac{1}{2}})

d er i v a t i v e cos^{4} (t) - sin^{4} (t)

\frac{d}{d x} (- \frac{a ^{2} x}{b ^{2} y})

\int_{0}^{9} π y^{2} d y