limit as x approaches infinity of sqrt((9x^3+8x-4)/(3-5x+x^3))

解答

x \to \infty lim (\frac{9 x ^{3} + 8 x - 4}{3 - 5 x + x ^{3}})

3

求解步骤

x \to \infty lim (\frac{9 x ^{3} + 8 x - 4}{3 - 5 x + x ^{3}})

除以分母的最高次幂:

\frac{9 + \frac{8}{x ^{2}} - \frac{4}{x ^{3}}}{\frac{3}{x ^{3}} - \frac{5}{x ^{2}} + 1}

= x \to \infty lim (\frac{9 + \frac{8}{x ^{2}} - \frac{4}{x ^{3}}}{\frac{3}{x ^{3}} - \frac{5}{x ^{2}} + 1})

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

不定型除外

= x \to \infty lim (\frac{9 + \frac{8}{x ^{2}} - \frac{4}{x ^{3}}}{\frac{3}{x ^{3}} - \frac{5}{x ^{2}} + 1})

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} ( 9 + \frac{8}{x ^{2}} - \frac{4}{x ^{3}} )}{lim _{x \to \infty} ( \frac{3}{x ^{3}} - \frac{5}{x ^{2}} + 1 )}

x \to \infty lim (9 + \frac{8}{x ^{2}} - \frac{4}{x ^{3}}) = 9

x \to \infty lim (\frac{3}{x ^{3}} - \frac{5}{x ^{2}} + 1) = 1

= \frac{9}{1}

化简

\frac{9}{1} : 3

= 3

t \to \infty lim (- 654 e^{- \frac{7 t}{2180}} + 654)

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