L'Hopital's Rule

Question

lim x²lnx

x→0⁺

Please explain each step in detail. Thank you!

Answer 1

x²lnx = lnx/1/x²

lim x->0+ lnx/1/x² = -∞/∞ So can use L'Hopital's rule

Take the derivative of the numerator and denominator

lim x->0+ (1/x)/(-2/x³) = lim x->0+ (1/x)(x³/-2) = lim x->0+ -x²/2

Taking the limit

lim x->0+ -x²/2 = -0²/2 = 0

Answer 2

x^2lnx = lnx/x^-2

(lnx)'/(x^-2)' = (x^-1)/(-2x^-3) = -x^2/2 = -0^2/2 = 0

the limit as x approaches 0+ of x^2lnx = 0

if you tried to plug 0 for x into x^2lnx it's 0 times negative infinity, an indeterminate form

I'd say try an online L'Hopital calculator, and that might work. Although I just tried 2 of them. One came back DNE, Does not exist. Another said the limit = 404. Not sure what happened.

For L'Hopital's Rule you need a fraction. x^2lnx can be converted into a fraction = lnx/x^-2

then take derivatives separately of numerator and denominator to get 1/x= x^-1 and -2x^-3

divide numerator by denominator to get [x^(-1+3)]/-2 = x^2/-2, now plug in 0 for x to get 0/-2 = 0= the limit

Another approach is let x= some very small number, like 0.0001

calculate x^2lnx= (.0001)^2ln.0001 and you get a far more smaller number -0.000000....

it appears to be approaching zero