f(x)= x^2+3 maka lim x mendekati 0 f(2+p) - f(2)÷ p =

f(2+p) = (2+p)²+3= 4+4p+p²+4
=8+4p+p²
f(2)=2³+3=8+3=11

Lim 8+4p+p²+11 / p
x→0
Lim 19+4p+p² / p
x→0
Lim p(19/p + 4 + p) / p
x→0
19/0 + 4 + 0
~ + 4 + 0
4
jadi hasil akhirnya 4