That's the derivative at 2 so is going to be 2(2)+1=5, second choice, but let's do it the hard way.
[tex]\displaystyle \lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h} = \lim_{h \to 0} \dfrac{(2+h)^2 + (2+h) + 1 -(2^2 + 2 + 1)}{h}[/tex]
[tex]\displaystyle =\lim_{h \to 0} \dfrac{4 + 4h +h^2 +h-4}{h}=\lim_{h \to 0} \dfrac{h^2 +5h}{h} = \lim_{h \to 0} h+5 = 5[/tex]
Answer: 5, second choice
