![\displaystyle\lim_(h\to0)\frac{f(9+h)-f(9)}h = \lim_(h\to0)\frac{(9+h)^4-9^4}h](https://img.qammunity.org/2022/formulas/mathematics/college/iynzqsb4sm3s9ee7gpwt8v9a489mn0rm71.png)
Carry out the binomial expansion in the numerator:
![(9+h)^4 = 9^4+4*9^3h+6*9^2h^2+4*9h^3+h^4](https://img.qammunity.org/2022/formulas/mathematics/college/upyiwmzhp50zbpyscxkeff6q2qvstd9918.png)
Then the 9⁴ terms cancel each other, so in the limit we have
![\displaystyle \lim_(h\to0)\frac{4*9^3h+6*9^2h^2+4*9h^3+h^4}h](https://img.qammunity.org/2022/formulas/mathematics/college/o9uli5yzxzremw335ju39k41pp6r9lzd2t.png)
Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:
![\displaystyle \lim_(h\to0)(4*9^3+6*9^2h+4*9h^2+h^3)](https://img.qammunity.org/2022/formulas/mathematics/college/db5qoyby4bta8k70duox9b7fjxx4re2wge.png)
Then when h converges to 0, each remaining term containing h goes to 0, leaving you with
![\displaystyle\lim_(h\to0)\frac{f(9+h)-f(9)}h = 4*9^3 = \boxed{2916}](https://img.qammunity.org/2022/formulas/mathematics/college/a30kwly859242cnmjjydqk2p8teypwzf8a.png)
or choice C.
Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:
![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](https://img.qammunity.org/2022/formulas/mathematics/college/r9kzdhc1xpwb0emmara25i0t21zxn0q7r0.png)
We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.