Find the limit as x approaches 1 (1/lnx - 1/x-1 ) using L\'Hopital\'s Rule. Find

Question

Find the limit as x approaches 1 (1/lnx - 1/x-1 ) using L'Hopital's Rule. Find the limit as x approaches 0 from the left (sinx lnx ) using L'Hopital's Rule. Find the limit as x approaches infinity (1+ (a/x) ) ^ bx using L'Hopital's Rule. Prove cosh-sinhx = e^-x

Explanation / Answer

lim x-->1 (1/lnx - 1/x-1 ) = lim x-->1 ((x - lnx - xlnx)/(xlnx)) apply L'Hopital's Rule = lim x-->1 ((1 - 1/x - 1 - lnx)/(1 + lnx) = (-1 - 0)/(1 + 0) = -1 lim x-->0 (sinx lnx ) apply L'Hopital's Rule = lim x-->0 (cosx lnx + sinx /x) = 0 + 1 = 1 lim x-->infy (1+ (a/x) ) ^ bx we have lim n -->infy (1 + 1/n)^n = e here let a/x = 1/n ==> x = an so, lim x-->infy (1+ (a/x) ) ^ bx = lim n-->infy ( 1 + 1/n)^(abn) = e^(ab) cosh-sinhx = [e^x + e^-x]/2 - [e^x - e^-x]/2 = [2e^-x]/2 = e^-x

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