Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim (6x – In(x)) XStep 1 As x → [infinity], In(x) Step 2 Therefore, lim (6x - In(x)) is indeterminate of type [infinity] – [infinity] . We will change x → [infinity] it to a product by factoring out 4x to get lim x → [infinity] 6x 6x(1 - In(x)/6xStep 3 We now have lim x → [infinity] (6x − ln(x)) = lim x → [infinity] 6x 1 − ln(x) 6x. Let's first focus on lim x → [infinity] ln(x)/6x . 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)/6x = lim x → [infinity] 1/6x = 0Step 4 Since lim x → [infinity] In(x)/4x = 0, then we can conclude that: x → [infinity] lim (6x - In(x)) = lim 6x(1 - In(x)/6x)