Evaluating limit with 2 unknown parameters I have another question involving limits with unknown parameters:
In this one, the limit approaches infinity:
$\lim \limits_{x \to 1 } {{\frac{x^2+ax+b}{(x-1)^2}}}= \infty $
This one is really causing me hardship, the only way i can see is to guess a function that would go to plus infinity that has a similar structure to this and choose the values of a and b, this way.
I was hoping there is another approach, some principle that can be used to figure this infinite limit out.
 A: Since evaluating at inputs close to 1 will produce an extremely small, but $\pmb{positive}$ denominator, we look at the numerator.  
Case 1;  $1+a+b=0$
First, Replacing $x$ with 1 gives
$$1+a+b$$
Now, if $a=-2, b=1$, then $1+a+b=0$, the numerator and denominator are the same and the limit is 1.  The function doesn't exist at $x=1$, but does for all other $x$.
However, if there is an $a$ and $b$ such that the numerator is a product of $x-1$ and some other linear term, what happens?  Then the denominator acts like $x-1$ and not like $(x-1)^2$  This occurs, for example, when $a=-3$ and $b=2$.  If you divide $x^2+ax+b$ by $x-1$ you get
$$x^2+ax+b=(x-1)(x+a+1)+(a+b+1)$$
But since $a+b+1=0$, there is no remainder and your limit changes to 
$$\lim_{x\rightarrow 1}\frac{x+1+a}{x-1}=\lim_{x\rightarrow 1}\left(1+\frac{a+2}{x-1}\right)$$
Thus, you can reduce the problem to finding the limit here:
$$\lim_{x\rightarrow 1}\frac{a+2}{x-1}=(a+2)\lim_{x\rightarrow 1}\frac{1}{x-1}$$
and this limit doesn't exist as left and right limits diverge.
Case 2: $1+a+b>0$
This is simple because then the numerator does not contain a factor of $x-1$ and since the numerator and denominator are positive when $x$ approaches 1, the limit increases without bound, i.e., $\infty$.
Case 3:  $1+a+b<0$
Same as case 2, but now the limit decreases without bound, $-\infty$.
