what is the limit x->1+ (x^3-1)ln(x-1)^5

Answer 1

I assume you mean limit_(x → 1+) [(x^3-1) ln((x-1)^5)], that is, you're taking the limit from the right as x goes to 1 of the function (x^3-1) ln((x-1)^5). The answer to this is 0. Let me sketch a solution.

First, note that

(x³-1) ln((x-1)⁵) = 5 (x³-1) ln(x-1) = 5 (x²+x + 1) (x-1) ln(x-1).

Here I took the exponent 5 out of the log and factored (x³-1).

Now note that 5 (x²+x + 1) → 15 as x → 1+, so using the product rule for limits, we only need to figure out the limit of (x-1) ln(x-1) as x → 1+. We can do this by applying L'Hôpital's rule to

(x-1) ln(x-1) = ln(x-1) / (1 / (x-1)).

The derivative of the numerator ln(x-1) is 1 / (x-1), and the derivative of the denominator 1 / (x-1) is - 1 / (x-1)². When you simplify and take the limit, you get 0.

Please let me know if you have any questions.