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(C) Given the limit $L = \lim_{h \rightarrow 0} \frac{f(2+2h+h^2)-f(2)}{f(1+h-h^2)-f(1)}$.Since the limit is in the $\frac{0}{0}$ form, we apply $L$'$

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equal to $3$

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(C) Given the limit $L = \lim_{h \rightarrow 0} \frac{f(2+2h+h^2)-f(2)}{f(1+h-h^2)-f(1)}$.Since the limit is in the $\frac{0}{0}$ form, we apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $h$:$L = \lim_{h \rightarrow 0} \frac{f^{\prime}(2+2h+h^2) \cdot (2+2h)}{f^{\prime}(1+h-h^2) \cdot (1-2h)}$.Now, substitute $h=0$ into the expression:$L = \frac{f^{\prime}(2+0+0) \cdot (2+0)}{f^{\prime}(1+0-0) \cdot (1-0)}$.$L = \frac{f^{\prime}(2) \cdot 2}{f^{\prime}(1) \cdot 1}$.Given $f^{\prime}(2)=6$ and $f^{\prime}(1)=4$, we have:$L = \frac{6 \cdot 2}{4 \cdot 1} = \frac{12}{4} = 3$.

$f(x)$ is a differentiable function and given $f^{\prime}(2)=6$ and $f^{\prime}(1)=4$, then $L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}$

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