limit as x approaches-infinity of (2x)/(\sqrt[3]{x^6-1)}

解答

x \to - \infty lim (\frac{2 x}{3 x ^{6} - 1})

0

求解步骤

x \to - \infty lim (\frac{2 x}{3 x ^{6} - 1})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 \cdot x \to - \infty lim (\frac{x}{3 x ^{6} - 1})

除以分母的最高次幂:

\frac{\frac{1}{x}}{3 1 - \frac{1}{x ^{6}}}

= 2 \cdot x \to - \infty lim \frac{\frac{1}{x}}{3 1 - \frac{1}{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

= 2 \cdot \frac{lim _{x \to - \infty} ( \frac{1}{x} )}{lim _{x \to - \infty} ( 3 1 - \frac{1}{x ^{6}} )}

x \to - \infty lim (\frac{1}{x}) = 0

x \to - \infty lim (3 1 - \frac{1}{x ^{6}}) = 1

= 2 \cdot \frac{0}{1}

化简

2 \cdot \frac{0}{1} : 0

= 0

x \to 0 lim (\frac{3 x + 1 + x - 1}{x})

x \to 0 lim (sin (\frac{3 x}{x}))

x \to 0 lim ((1 + \frac{x}{4})^{\frac{1}{x}})

x \to - 1 lim (x^{3} - 3 x + 2)

x \to 2 lim (\frac{x - 1}{x + 1})