What is the limit as n approaches infinity of n^{-1} ?

The limit as n approaches infinity of n^{-1} is 0

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limit as n approaches infinity of n^{-1}

n \to \infty lim (n^{- 1})

0

Solution steps

n \to \infty lim (n^{- 1})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= (n \to \infty lim (n))^{- 1}

Apply the common limit:

n \to \infty lim (n) = \infty

= \infty^{- 1}

Apply Infinity Property:

\infty^{- c} = 0

= 0

y^{^{''}} - y^{^{'}} - 3 y = 0

f (x) = ln ∣ cot (3 \cdot x) ∣

((3 x^{2} + 3 x + 4)^{^{'}} (3 x^{2} + 6 x)^{^{'}})^{^{'}}

\int e^{t a n (2 x)} sec^{2} (2 x) d x

y = x^{3} - 2 x^{2} + 5, (2, 5)

What is the limit as n approaches infinity of n^{-1} ?
The limit as n approaches infinity of n^{-1} is 0