limit as x approaches infinity of 2^{x+1}3^{-x}

解答

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

0

求解步骤

x \to \infty lim (2^{x + 1} \cdot 3^{- x})

化简

2^{x + 1} : 2^{x} \cdot 2

= x \to \infty lim (2^{x} \cdot 2 \cdot 3^{- x})

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

= 2 \cdot x \to \infty lim (2^{x} \cdot 3^{- x})

化简

3^{- x} : \frac{1}{3 ^{x}}

= 2 \cdot x \to \infty lim (2^{x} \cdot \frac{1}{3 ^{x}})

化简

2^{x} \cdot \frac{1}{3 ^{x}} : (\frac{2}{3})^{x}

= 2 \cdot x \to \infty lim ((\frac{2}{3})^{x})

运用一般极限法则:

x \to \infty lim (a^{x}) = 0, 0 < a < 1

= 2 \cdot 0

化简

= 0

x \to - 10 - lim (\frac{1}{( x + 10 ) ^{6}})

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n \to \infty lim (n)