limit as x approaches infinity of sqrt(4x^2-7x)-2x

解答

x \to \infty lim (4 x^{2} - 7 x - 2 x)

- \frac{7}{4}

十进制

- 1.75

求解步骤

x \to \infty lim (4 x^{2} - 7 x - 2 x)

乘以

4 x^{2} - 7 x - 2 x

的共轭根式

: \frac{- 7 x}{4 x ^{2} - 7 x + 2 x}

= x \to \infty lim (\frac{- 7 x}{4 x ^{2} - 7 x + 2 x})

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

= - 7 \cdot x \to \infty lim (\frac{x}{4 x ^{2} - 7 x + 2 x})

除以分母的最高次幂:

\frac{1}{4 - \frac{7}{x} + 2}

= - 7 \cdot x \to \infty lim \frac{1}{4 - \frac{7}{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

不定型除外

= - 7 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 4 - \frac{7}{x} + 2 )}

x \to \infty lim (1) = 1

x \to \infty lim (4 - \frac{7}{x} + 2) = 4

= - 7 \cdot \frac{1}{4}

化简

- 7 \cdot \frac{1}{4} : - \frac{7}{4}

= - \frac{7}{4}

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