Computing the limit directly:
[tex]\displaystyle\lim_{h\to0}\frac{f(2+h)-f(2)}h=\lim_{h\to0}\frac{((2+h)^2+(2+h)+1)-7}h\\\displaystyle=\lim_{h\to0}\frac{5h+h^2}h\\=\displaystyle\lim_{h\to0}(5+h)=\boxed{5}[/tex]
Alternatively, you can recognize the limit as being equivalent the derivative of f(x) at x = 2, in which case differentiating and plugging in 2 gives
f'(x) = 2x + 1 => f' (2) = 5
