limit as k approaches infinity of (k^{1/k})/(ln(k))

解答

k \to \infty lim (\frac{k ^{\frac{1}{k}}}{ln ( k )})

0

求解步骤

k \to \infty lim (\frac{k ^{\frac{1}{k}}}{ln ( k )})

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 _{k \to \infty} ( k ^{\frac{1}{k}} )}{lim _{k \to \infty} ( ln ( k ) )}

k \to \infty lim (k^{\frac{1}{k}}) = 1

k \to \infty lim (ln (k)) = \infty

= \frac{1}{\infty}

运用无穷大属性:

\frac{c}{\infty} = 0

= 0

\frac{d}{d s} (\frac{s ^{2} - 4 s + 20}{( s + 2 ) ( s + 4 )})

\int e^{t a n (6 x)} sec^{2} (6 x) d x

y^{^{'}} = 6 x^{2} y

im pl i c i t \frac{d y}{d x}, 4 x = y^{2}

d er i v a t i v e \frac{e ^{x}}{3 x ^{2} + 7}