Let f(x) be a polynomial function such that $f\left(x\right)+f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)=x^5+64$. Then, the value of $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{x-1}$

$$\begin{gathered} {Lt}_{x\to 1}\frac{f\left(x\right)}{x-1}=f^{\text{'}}\left(1\right) \left(\text{ and }f\right(1)=0), \\ given that f\left(x\right)+f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)=x^5+64 \\ \Rightarrow f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)+f^{\text{'}\text{'}\text{'}}\left(x\right)=5x^4 \\ \Rightarrow f^{\text{'}\text{'}}\left(x\right)+f^{\text{'}\text{'}\text{'}}\left(x\right)+f^{iv}\left(x\right)=20x^3 \\ \Rightarrow f^{\text{'}\text{'}\text{'}}\left(x\right)+f^{iv}\left(x\right)+f^v\left(x\right)=60x^2 \\ \therefore f^v\left(x\right)-f^{\text{'}\text{'}}\left(x\right)=60x^2-20x^3 \\ \Rightarrow 120-{\mathrm{f}}^{\text{'}\text{'}}\left(1\right)=40\Rightarrow {\mathrm{f}}^{\text{'}\text{'}}\left(1\right)=80 \\ \text{ Also }f\left(1\right)+f^{\text{'}}\left(1\right)+{\mathrm{f}}^{\text{'}\text{'}}\left(1\right)=65\Rightarrow f^{\text{'}}\left(1\right)=-15 \end{gathered}$$