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

Answer 1

corrrected

The indeterminate form of the limit is 0*-∞ as written

rewrite: L=lim x->1⁺ of [1/(1/(x³-1))]*5(ln(x-1))

now L is of form (-)∞/∞

and can proceed to use L'Hopital once

now L=lim x->1⁺ of [5/(x-1)]/[[3x²/(x³-1)²]=[5(x³-1)²]/[3x²(x-1)]

which now has indeterminate form 0/0, so can apply L'Hopital again

now L=limx->1⁺ of [10(x³-1)3x²]/[3x²+6x(x-1)]

L=0/3=0