Using L’hopital’s rule please I need this ASAP Is due today!!!!!!!!!

Answer:See below.Step-by-step explanation:You differentiate  top and bottom of the fraction until substitution gives you a value.I can do the third one for you: Lim x --> 0 of sin2x / sin3x= lim x --> 0 of 2 cos2x / 3 cos 3x= 2 cos 0 / 3 cos 0= 2/3.Limit as x--> 0  of (e^x - (1 - x) / x= limit as x --> 0 of e^x + x - 1 / x=  lim (e^x + 1) / 1= 1 + 1 / 1= 2.limit as x-->  00  of  3x^2 - 2x +  1/ (2x^2 + 3)= limit as x --> 00 of  6x - 2 / 4x     ( 00 = infinity)Applying l'hopitals rule again:limit is 6 / 4 =  3/2.Limit  as x --> 00 of (ln x)^3 / x= limit  3 (Ln x)^2 ) / x= limit of  6 ln x / x= limit  6 / x= 0.We had  to apply l'hopitals  rule 3 times here,