lim-x-x-ln-e-x-1-e-x-1-

Question Number 195453 by Rodier97 last updated on 02/Aug/23

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

Commented by Frix last updated on 02/Aug/23

0

Answered by MM42 last updated on 02/Aug/23

{l i m}_{x \to \infty} l n {(\frac{e^{x} + 1}{e^{x} - 1})}^{x} ★

A = {(\frac{e^{x} + 1}{e^{x} - 1})}^{x} \Rightarrow {l i m}_{x \to \infty} A = e^{{l i m}_{x \to \infty} (\frac{e^{x} + 1}{e^{x} - 1} - 1) x}

= e^{{l i m}_{x \to \infty} (\frac{e^{x} + 1}{e^{x} - 1} - 1) x} = e^{{l i m}_{x \to \infty} (\frac{2 x}{e^{x} - 1})} = e^{0} = 1

\Rightarrow ★ = l n 1 = 0 ✓

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