limit as x approaches infinity of (3^{x-3})/(2^{n+3)}

解答

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

\infty

求解步骤

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

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

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

化简

3^{x - 3} : 3^{x} \cdot 3^{- 3}

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

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

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

3^{- 3} = \frac{1}{27}

= \frac{1}{2 ^{n + 3}} \cdot \frac{1}{27} \cdot x \to \infty lim (3^{x})

运用一般极限法则:

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

= \frac{1}{2 ^{n + 3}} \cdot \frac{1}{27} \cdot \infty

化简

\frac{1}{2 ^{n + 3}} \cdot \frac{1}{27} \cdot \infty : \infty

= \infty

x \to 2 - lim (x + 4)

3 y (x^{2} - 1) d x + (x^{3} + 8 y - 3 x) d y = 0

s l o p e y = x^{2} - 5

d er i v a t i v e f (x) = \frac{1}{2} x - \frac{1}{3}

d er i v a t i v e \frac{470 t}{( t ^{2} + 6 ) ^{3}}