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

解

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

0

解答ステップ

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

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

不定形式の場合を除く

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

最高の分母べき乗で割る:

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

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

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

不定形式の場合を除く

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

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

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

= 3 \frac{0}{1}

簡素化

3 \frac{0}{1} : 0

= 0