limit as x approaches-infinity of (\sqrt[4]{x^4-3}+6x)/(1-2x)

解

x \to - \infty lim (\frac{4 x ^{4} - 3 + 6 x}{1 - 2 x})

- \frac{5}{2}

十進法表記

- 2.5

解答ステップ

x \to - \infty lim (\frac{4 x ^{4} - 3 + 6 x}{1 - 2 x})

最高の分母べき乗で割る:

\frac{- 4 1 - \frac{3}{x ^{4}} + 6}{\frac{1}{x} - 2}

= x \to - \infty lim \frac{- 4 1 - \frac{3}{x ^{4}} + 6}{\frac{1}{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

= \frac{lim _{x \to - \infty} ( - 4 1 - \frac{3}{x ^{4}} + 6 )}{lim _{x \to - \infty} ( \frac{1}{x} - 2 )}

x \to - \infty lim (- 4 1 - \frac{3}{x ^{4}} + 6) = 5

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

= \frac{5}{- 2}

簡素化

= - \frac{5}{2}