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lim x->1 (x/x-1 - 1/ln(x))
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use l'hospital's rule where appropriate
lhospitals rule u just take derivative of top and bottom separately
lim (x -> 1) [x/(x - 1) - 1/lnx] lim (x -> 1) [(xlnx - (x-1)] / [(x-1)lnx] Using L'Hospital's rule, limit lim (x -> 1) (1 + lnx - 1) / [lnx + (x-1)/x] lim (x -> 1) (1/x) / (1/x + 1/x^2) lim (x -> 1) 1 / (1 + 1/x) 1/2.
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