limit as x approaches infinity of (3e^x)/(9+2e^{5x)}

解答

x \to \infty lim (\frac{3 e ^{x}}{9 + 2 e ^{5 x}})

0

求解步骤

x \to \infty lim (\frac{3 e ^{x}}{9 + 2 e ^{5 x}})

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

= 3 \cdot x \to \infty lim (\frac{e ^{x}}{9 + 2 e ^{5 x}})

Divide by

e^{5 x} : \frac{\frac{1}{e ^{4 x}}}{\frac{9}{e ^{5 x}} + 2}

= 3 \cdot x \to \infty lim (\frac{\frac{1}{e ^{4 x}}}{\frac{9}{e ^{5 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

= 3 \cdot \frac{lim _{x \to \infty} ( \frac{1}{e ^{4 x}} )}{lim _{x \to \infty} ( \frac{9}{e ^{5 x}} + 2 )}

x \to \infty lim (\frac{1}{e ^{4 x}}) = 0

x \to \infty lim (\frac{9}{e ^{5 x}} + 2) = 2

= 3 \cdot \frac{0}{2}

化简

3 \cdot \frac{0}{2} : 0

= 0

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

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