Mathematics, 29.11.2019 05:31 bay62
We now have lim x → [infinity] (4x − ln(x)) = lim x → [infinity] 4x 1 − ln(x) 4x . let's first focus on lim x → [infinity] ln(x) 4x . since ln(x) → [infinity] as x → [infinity], then this limit is indeterminate of type [infinity]/[infinity]. using l'hospital's rule, we find: lim x → [infinity] ln(x) 4x = lim x → [infinity] 1 = 0 .
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We now have lim x → [infinity] (4x − ln(x)) = lim x → [infinity] 4x 1 − ln(x) 4x . let's first focus...
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