What is the limit as n approaches infinity of 2^{1/n} ?

The limit as n approaches infinity of 2^{1/n} is 1

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limit as n approaches infinity of 2^{1/n}

n \to \infty lim (2^{\frac{1}{n}})

1

Solution steps

n \to \infty lim (2^{\frac{1}{n}})

Apply exponent rules

= n \to \infty lim (e^{\frac{1}{n} l n (2)})

Apply the Limit Chain Rule:

1

= 1

y^{^{'}} + e^{x} y = e^{3 x} y^{- 2}

6 - x^{2}, x^{2} - 2

\int \frac{x}{x + 1} d x

\int_{- 2}^{2} ln (x + 3) d x

\int 2 x e^{17 x} d x

What is the limit as n approaches infinity of 2^{1/n} ?
The limit as n approaches infinity of 2^{1/n} is 1