limit as x approaches infinity of 1/(6^{x+1)}

解答

x \to \infty lim (\frac{1}{6 ^{x + 1}})

0

求解步骤

x \to \infty lim (\frac{1}{6 ^{x + 1}})

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} ( 1 )}{lim _{x \to \infty} ( 6 ^{x + 1} )}

x \to \infty lim (1) = 1

x \to \infty lim (6^{x + 1}) = \infty

= \frac{1}{\infty}

运用无穷大属性:

\frac{c}{\infty} = 0

= 0

s l o p e (6, 2), (- 3, 5)

d er i v a t i v e \frac{2 x - 3}{4 x + 1}

\int (3 x - 1) x + 6 d x

d er i v a t i v e ln (\frac{2 x + 1}{x ^{2} - 4})

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