Find $a, b\in\mathbb{R}$ so that $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite. 
Find $a, b\in\mathbb{R}$ so that 
  $$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$$
   is finite. 

I tried to use polynomial division, but the computations get tedious really fast. Any suggestions? 

First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.
