Limit of logarithms exponential lim_(x->oo)((ln(x-2))/(ln(x-1)))^(x\ln x). L'Hopital seems like a very hardcore solutions given the situation.Are the any other options? reevelingw97
Answered question 2022-11-21
lim x → ∞ ( ln ( x − 2 ) ln ( x − 1 ) ) x ln x . L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?
Answer & Explanation Let f ( x ) = ( ln ( x − 2 ) ln ( x − 1 ) ) x ln x . Then ln f ( x ) = x ln x [ ln ln ( x − 2 ) − ln ln ( x − 1 ) ] = x ln x [ ln ( 1 + ln ( 1 − 2 / x ) ln x ) − ln ( 1 + ln ( 1 − 1 / x ) ln x ) ] = x ln x [ ln ( 1 − 2 x ln x + o ( 1 x ln x ) ) − ln ( 1 − 1 x ln x + o ( 1 x ln x ) ) ] = x ln x [ − 2 x ln x + 1 x ln x + o ( 1 x ln x ) ] = − 1 + o ( 1 ) . Therefore lim x → + ∞ f ( x ) = e − 1
Let L be the desired limit so that log L = log { lim x → ∞ ( log ( x − 2 ) log ( x − 1 ) ) x log x } = lim x → ∞ log ( log ( x − 2 ) log ( x − 1 ) ) x log x (by continuity of log) = lim x → ∞ x log x log ( log ( x − 2 ) log ( x − 1 ) ) = − lim t → 0 + log t t log ( log ( 1 − 2 t ) − log t log ( 1 − t ) − log t ) (putting x = 1 / t ) = − lim t → 0 + log t t log ( 1 − log ( 1 − t ) − log ( 1 − 2 t ) log ( 1 − t ) − log t ) = − lim t → 0 + log t t log ( 1 − log ( 1 − t ) − log ( 1 − 2 t ) log ( 1 − t ) − log t ) − log ( 1 − t ) − log ( 1 − 2 t ) log ( 1 − t ) − log t ( − log ( 1 − t ) − log ( 1 − 2 t ) log ( 1 − t ) − log t ) = lim t → 0 + log t t log ( 1 − t ) − log ( 1 − 2 t ) log ( 1 − t ) − log t = lim t → 0 + log ( 1 − t ) − log ( 1 − 2 t ) t 1 log ( 1 − t ) log t − 1 = lim t → 0 + log ( 1 − t ) − log ( 1 − 2 t ) t ⋅ 1 0 − 1 = − lim t → 0 + log ( 1 − t 1 − 2 t ) t = − lim t → 0 + log ( 1 + t 1 − 2 t ) t = − lim t → 0 + log ( 1 + t 1 − 2 t ) t 1 − 2 t ⋅ 1 1 − 2 t = − lim t → 0 + 1 1 − 2 t = − 1 and hence L = 1 / e .
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