Using l’hopital’s rule i need this asap is due !

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,