Respuesta :

gerishonwambugu
-3 is the answer use L hospital's rule.
mreijepatmat
lim (x²-3sinx)/3 => lim(x²/x) -lim(3sinx)/x
x→0
lim(x²/x) -lim(3sinx)/x ==> lim(x) -lim (3sinx)x
x→0         x→0
= 0 + lim(3sinx/x)
Let's find the limite of 3(sinx/x)
Limit os sinx/x when x →0 equal to 1 & for 3sinx it will be 3

So 〖lim〗┬(x→0) (x^2-3sinx)/x  = 3