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,
