limit as x approaches infinity of 3x-sqrt(9x^2+x)

解答

x \to \infty lim (3 x - 9 x^{2} + x)

- \frac{1}{6}

十进制

- 0.16666 \dots

求解步骤

x \to \infty lim (3 x - 9 x^{2} + x)

乘以

3 x - 9 x^{2} + x

的共轭根式

: \frac{- x}{3 x + 9 x ^{2} + x}

= x \to \infty lim (\frac{- x}{3 x + 9 x ^{2} + x})

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

= - x \to \infty lim (\frac{x}{3 x + 9 x ^{2} + x})

除以分母的最高次幂:

\frac{1}{3 + 9 + \frac{1}{x}}

= - x \to \infty lim \frac{1}{3 + 9 + \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

不定型除外

= - \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 3 + 9 + \frac{1}{x} )}

x \to \infty lim (1) = 1

x \to \infty lim (3 + 9 + \frac{1}{x}) = 6

= - \frac{1}{6}

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