What is the limit as n approaches infinity of 2^{-n} ?

The limit as n approaches infinity of 2^{-n} is 0

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limit as n approaches infinity of 2^{-n}

n \to \infty lim (2^{- n})

0

Solution steps

n \to \infty lim (2^{- n})

Apply exponent rules

= n \to \infty lim (e^{- n l n (2)})

Apply the Limit Chain Rule:

0

= 0

\int_{0}^{- 0.5} 70 x d x

\frac{\partial}{\partial x} (x \cdot sin (y))

\frac{\partial}{\partial x} (y - y x^{2})

f (x) = \frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

\frac{\partial}{\partial x} (- 2 k l (ln (\frac{y}{- x - y})))

What is the limit as n approaches infinity of 2^{-n} ?
The limit as n approaches infinity of 2^{-n} is 0