lim-x-x-1-x-where-is-a-fractional-part-of-x-

Question Number 205174 by universe last updated on 12/Mar/24

\lim_{x \to \infty} {x^{1 / x}} = ? w h e r e {.} i s a f r a c t i o n a l p a r t o f x

Answered by lepuissantcedricjunior last updated on 12/Mar/24

\lim_{x \to + \infty} {(x)}^{\frac{1}{x}} = \infty^{0} = F I

\lim_{x \to + \infty} {(x)}^{\frac{1}{x}} = \lim_{x \to + \infty} e^{\frac{l n x}{x}} = e^{0} = 1

c a r \lim_{x \to + \infty} \frac{l n x}{x} = 0 f o r m u l e u s u e l l e

\dots \dots \dots l e p u i s s a n t D r \dots \dots \dots \dots

Answered by Berbere last updated on 13/Mar/24

= x^{\frac{1}{x}} - [x^{\frac{1}{x}}]

x \overset{f}{\to} x^{\frac{1}{x}};

f (x) = e^{\frac{1}{x} l n (x)}; \lim_{x \to \infty} f (x) = 1

x > 1 \Rightarrow f (x) ⩾ 1 \Rightarrow \exists A \in R, \forall x ⩾ A \Rightarrow 1 ⩽ f (x) ⩽ \frac{3}{2}

\Rightarrow \forall x ⩾ A [e^{\frac{l n (x)}{x}}] = 1

\Rightarrow x^{\frac{1}{x}} - [x^{\frac{1}{x}}] = x^{\frac{1}{x}} - 1; \forall x ⩾ A

\lim_{x \to \infty} x^{\frac{1}{x}} - 1 = 0

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