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Evaluate lim x β†’ 0+ x ln(x3). solution the given limit is indeterminate because, as x β†’ 0+, the first factor (x) approaches 0 correct: your answer is correct. while the second factor ln(x3) approaches βˆ’βˆž. writing x = 1/(1/x), we have 1/x β†’ ∞ as x β†’ 0+, so l'hospital's rule gives lim x β†’ 0+ x ln(x3) = lim x β†’ 0+ ln(x3) 1/x = lim x β†’ 0+ 3/x βˆ’1/x2 = lim x β†’ 0+ incorrect: your answer is incorrect. = .

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