What is the limit as n approaches infinity of e^{1/n ln(n)} ?

The limit as n approaches infinity of e^{1/n ln(n)} is 1

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limit as n approaches infinity of e^{1/n ln(n)}

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

1

Solution steps

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

Simplify

e^{\frac{1}{n} l n (n)} : n^{\frac{1}{n}}

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

Apply exponent rules

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

Apply the Limit Chain Rule:

1

= 1

\frac{d}{d x} (2 x^{3} - 3)

y^{^{'}} = \frac{y}{x}, y (1) = 2

\frac{d}{d x} (\frac{2 x ^{2} - 2 x + 2}{( 2 x - 1 ) ^{2}})

(4.5) (9.25)

x \to - \infty lim (\frac{5}{x})

What is the limit as n approaches infinity of e^{1/n ln(n)} ?
The limit as n approaches infinity of e^{1/n ln(n)} is 1