anonymous
  • anonymous
Evaluate lim (cosh^-1(x)-ln(x)) as x goes to infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dan815
  • dan815
arccosh or 1/cosh(x)?
anonymous
  • anonymous
arccosh
dan815
  • dan815
so remember cosh can be written in terms of e and ln is the inverse of e

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anonymous
  • anonymous
I don't remember can you write to me ? please
ganeshie8
  • ganeshie8
Let \(y=\arccos hx \) \(\implies x = \cosh y = \dfrac{e^y + e^{-y}}{2}\) \(\implies 2x = e^y+e^{-y}\) \(\implies e^{2y} -2xe^y +1=0\) this is a quadratic, solving should give you \(y=\arccos h x = \ln(x + \sqrt{x^2-1)}\)
ganeshie8
  • ganeshie8
plug that thing in the given expression rest should be easy
anonymous
  • anonymous
thank you so much.. it is helpful
ganeshie8
  • ganeshie8
np, here is the rest : \(\lim\limits_{x\to\infty} \cosh^{-1}(x)-\ln(x) \) \(=\lim\limits_{x\to\infty} \ln(x+\sqrt{x^2-1})-\ln(x) \) \(=\lim\limits_{x\to\infty} \ln(\dfrac{x+\sqrt{x^2-1}}{x}) \) \(=\lim\limits_{x\to\infty} \ln(1+\sqrt{1-\dfrac{1}{x^2}}) \) \(= \ln(\lim\limits_{x\to\infty} 1+\sqrt{1-\dfrac{1}{x^2}}) \) \(=\ln(1+1)\)
anonymous
  • anonymous
Thank you soooooooooooo much
anonymous
  • anonymous
ln(1+1)

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