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Question Number 178412 by Spillover last updated on 16/Oct/22

$$\mathrm{Express}\sinh {}^{-1}\mathrm{x}-\ln \mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}$$$$\mathrm{natural}\mathrm{logarithms}.\mathrm{Hence}\mathrm{find}$$$$\mathrm{the}\mathrm{limit}\mathrm{as}\mathrm{x}\to \mathrm{\infty }$$

Answered by Ar Brandon last updated on 16/Oct/22

$${\sinh }^{-1}x-\ln x=\ln (x+\sqrt{1+x^2})-\ln x$$$$\underset{x\to \mathrm{\infty }}{\lim }=\underset{x\to \mathrm{\infty }}{\lim ln}\left(\frac{x+\sqrt{1+x^2}}{x}\right)$$$$=\underset{x\to \mathrm{\infty }}{\lim ln}(1+\sqrt{\frac{1}{x^2}+1})=\ln 2$$

Commented by Spillover last updated on 16/Oct/22

$$thanks$$

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