limit as x approaches-infinity of (sqrt(x^2+8x))/(8x+1)

解

x \to - \infty lim (\frac{x ^{2} + 8 x}{8 x + 1})

- \frac{1}{8}

十進法表記

- 0.125

解答ステップ

x \to - \infty lim (\frac{x ^{2} + 8 x}{8 x + 1})

最高の分母べき乗で割る:

\frac{- 1 + \frac{8}{x}}{8 + \frac{1}{x}}

= x \to - \infty lim \frac{- 1 + \frac{8}{x}}{8 + \frac{1}{x}}

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

= - x \to - \infty lim \frac{1 + \frac{8}{x}}{8 + \frac{1}{x}}

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} ( 1 + \frac{8}{x} )}{lim _{x \to - \infty} ( 8 + \frac{1}{x} )}

x \to - \infty lim (1 + \frac{8}{x}) = 1

x \to - \infty lim (8 + \frac{1}{x}) = 8

= - \frac{1}{8}