Question

Evaluate
lim (cosh^-1(x)-ln(x)) as x goes to infinity

Answer 1

arccosh or 1/cosh(x)?

Answer 2

arccosh

Answer 3

so remember cosh can be written in terms of e
and ln is the inverse of e

Answer 4

I don't remember can you write to me ? please

Answer 5

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
\(y=\arccos h x = \ln(x + \sqrt{x^2-1)}\)

Answer 6

plug that thing in the given expression
rest should be easy

Answer 7

thank you so much.. it is helpful

Answer 8

np, here is the rest :

\(\lim\limits_{x\to\infty} \cosh^{-1}(x)-\ln(x) \)

\(=\lim\limits_{x\to\infty} \ln(x+\sqrt{x^2-1})-\ln(x) \)

\(=\lim\limits_{x\to\infty} \ln(\dfrac{x+\sqrt{x^2-1}}{x}) \)

\(=\lim\limits_{x\to\infty} \ln(1+\sqrt{1-\dfrac{1}{x^2}}) \)

\(= \ln(\lim\limits_{x\to\infty} 1+\sqrt{1-\dfrac{1}{x^2}}) \)

\(=\ln(1+1)\)

Answer 9

Thank you soooooooooooo much

Answer 10

ln(1+1)