Evaluate the following limits using l’Hopital’s rule when needed: (A) lim x→∞ (ln x)^3/ 2x^2 (B) lim x→0+ sin(x)^(tan(x))

Answer 1

A)

lim x→∞ ln³(x)/2x²

This is a ∞/∞ case

Need to repeatedly use L'Hopital

3ln²(x)(1/x)/4x = (3/4)ln²(x)/x²

Again

(6/8)ln(x)(1/x)/x = (6/8)ln(x)/x²

Again

(6/16)(1/x)/x = (6/16)/x²

lim x→∞ 6/(16x²) = 0

B)

lim x→0+ sin(x)^tan(x)

sin(x)^tan(x) = exp(ln(sin(x)^tan(x))) = exp(tan(x)ln(sin(x))

Focus on lim x→0+ tan(x)ln(sin(x)) = lim x→0+ sin(x)ln(sin(x))/cos(x) = lim x→0+ sin(x)ln(sin(x))/1

let u = sin(x), lim u→0+ as lim x→0+

lim u→0+ uln(u) = lim u→0+ ln(u)/(1/u)

Now we can use L'Hopital's rule

lim u→0+ (1/u)/(-1/u²) = lim u→0+ -u = 0

exp(0) = 1