1. We are given that $f(x)+f'(x)+f''(x)=x^5+64$ and $f(x)$ is a polynomial.

We need to find $\lim_{x\to 1}\frac{f(x)}{x-1}.$

2. First, note that for this limit to be finite, we must have $f(1)=0,$ so let us check whether this is true from the given relation.

Substitute $x=1$: $f(1)+f'(1)+f''(1)=1^5+64=65.$ This alone does not directly give $f(1)$, so we find the polynomial explicitly.

3. Let $f(x)=ax^5+bx^4+cx^3+dx^2+ex+g.$ Then, $f'(x)=5ax^4+4bx^3+3cx^2+2dx+e,$ $f''(x)=20ax^3+12bx^2+6cx+2d.$

So, \begin{align*} f(x)+f'(x)+f''(x) &= ax^5 +(b+5a)x^4 +(c+4b+20a)x^3 \ &\quad +(d+3c+12b)x^2 +(e+2d+6c)x +(g+e+2d). \end{align*}

This must equal $x^5+64.$

4. Compare coefficients:

• Coefficient of $x^5$: $a=1$

• Coefficient of $x^4$: $b+5a=0 \implies b+5=0 \implies b=-5$

• Coefficient of $x^3$: $c+4b+20a=0 \implies c+4(-5)+20=0 \implies c=0$

• Coefficient of $x^2$: $d+3c+12b=0 \implies d+0+12(-5)=0 \implies d=60$

• Coefficient of $x$: $e+2d+6c=0 \implies e+120+0=0 \implies e=-120$

• Constant term: $g+e+2d=64 \implies g-120+120=64 \implies g=64$

Hence, $f(x)=x^5-5x^4+60x^2-120x+64.$

5. Now compute $f(1)$: $f(1)=1-5+60-120+64=0.$ So the limit becomes $\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1).$

6. Differentiate: $f'(x)=5x^4-20x^3+120x-120.$ Therefore, $f'(1)=5-20+120-120=-15.$

So, $\lim_{x\to 1}\frac{f(x)}{x-1}=-15.$

7. Comparing with the options, the correct option is: $\boxed{\text{A: }-15}$