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lim-x-ln-x-x-2-1-x-x-2-1-ln-2-x-1-x-1-




Question Number 193262 by josemate19 last updated on 09/Jun/23
lim_(x→+∞) (ln((x+(√(x^2 +1)))/(x+(√(x^2 −1)))).ln^2  ((x+1)/(x−1)))
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({ln}\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}.{ln}^{\mathrm{2}} \:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$ \\ $$
Answered by qaz last updated on 09/Jun/23
lim_(x→+∞) ((x+(√(x^2 +1)))/(x+(√(x^2 −1))))=lim_(x→+∞) ((x+1)/(x−1))=1  ⇒lim_(x→+∞) ln((x+(√(x^2 +1)))/(x+(√(x^2 −1))))ln^2 ((x+1)/(x−1))=0
$$\underset{{x}\rightarrow+\infty} {{lim}}\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\underset{{x}\rightarrow+\infty} {{lim}}\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow\underset{{x}\rightarrow+\infty} {{lim}ln}\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{ln}^{\mathrm{2}} \frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{0} \\ $$

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