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

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

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

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

\infty

Solution steps

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})

Simplify

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})

Apply the common limit:

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

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

Simplify

\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

y = x^{2} - 5

f (x) = \frac{1}{2} x - \frac{1}{3}

\frac{470 t}{( t ^{2} + 6 ) ^{3}}

What is the limit as x approaches infinity of (3^{x-3})/(2^{n+3)} ?
The limit as x approaches infinity of (3^{x-3})/(2^{n+3)} is infinity