limit as x approaches infinity of (4x^3+8)/(4x^3+sqrt(64x^6+3))

解答

x \to \infty lim (\frac{4 x ^{3} + 8}{4 x ^{3} + 64 x ^{6} + 3})

\frac{1}{3}

十进制

0.33333 \dots

求解步骤

x \to \infty lim (\frac{4 x ^{3} + 8}{4 x ^{3} + 64 x ^{6} + 3})

除以分母的最高次幂:

\frac{4 + \frac{8}{x ^{3}}}{4 + 64 + \frac{3}{x ^{6}}}

= x \to \infty lim \frac{4 + \frac{8}{x ^{3}}}{4 + 64 + \frac{3}{x ^{6}}}

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} ( 4 + \frac{8}{x ^{3}} )}{lim _{x \to \infty} ( 4 + 64 + \frac{3}{x ^{6}} )}

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

x \to \infty lim (4 + 64 + \frac{3}{x ^{6}}) = 12

= \frac{4}{12}

约分:

4

= \frac{1}{3}

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

x \to 1 lim (2 x + 2)

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

\int \frac{- 1}{x ^{2} + 16} d x

\frac{d}{d x} (\frac{x ^{2} - 3 x}{2 x + 1})