<s>[INST] 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? 
 [/INST] Hint:
Use the following lemma:

Lemma: $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite if and only if $(x-1)^2$ divides $ax^6+bx^5+1$.

Proof: If $ax^6+bx^5+1 = p(x)\cdot (x-1)^2$ then $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2} = p(1)$ is finite. For the other direction, if $(x-1)^2$ does not divide $ax^6+bx^5+1$ then one of the following occure
(1) $(x-1)$ does not divide $ax^6+bx^5+1$ in this case $x=1$ is not a root of $ax^6+bx^5+1$ and so the limit is infinite (unformally it equals to "$\frac{a+b+1}{0}$")
(2) $(x-1)$ divides $ax^6+bx^5+1$, let $p(x)$ be such that $ax^6+bx^5+1 = p(x)(x-1)$ then 
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2} = \lim_{x\rightarrow 1} \frac{p(x)}{x-1}$$ and $(x-1)$ doesn't divide $p(x)$ so as before the limit is infinite (unformally equals to "$\frac{p(1)}{0}$").
So your goal is to find all the $a,b$ such that $(x-1)^2$ divides $ax^6+bx^5+1$. This means that $x=1$ is a root of this polynomial and then once you divide by $(x-1)$ the integer $1$ is still a root of the polynomial. You will get two equations in the variables $a,b$ which are not hard to solve.
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