[tex]\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h[/tex]
Carry out the binomial expansion in the numerator:
[tex](9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4[/tex]
Then the 9⁴ terms cancel each other, so in the limit we have
[tex]\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h[/tex]
Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:
[tex]\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)[/tex]
Then when h converges to 0, each remaining term containing h goes to 0, leaving you with
[tex]\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}[/tex]
or choice C.
Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:
[tex]f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h[/tex]
We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.
