A calculus problem by Mehdi K.

Since the limit exists and is finite, therefore (i) $a\times 1^2+b\times 1+2=0$ or $a+b=\boxed {-2}$ , and (ii) $2a\times 1+b=2a+b=\boxed 1$ (since as $x$ tends to $1$ , $x-1$ tends to $0$ and the given fraction tends to $1$ ). Therefore $a=\boxed 3$ and $b=\boxed {-5}$ .

The limit of the denominator is 0, so for the limit of the fraction to be 1, the limit of the numerator must be 0 as well. This implies $a + b + 2 = 0$ , so $a + b = -2$ .

Rewrite the expression $((ax-2)(x-1)+k)/(x-1)$ ,

ie factorise the numerator by $x-1$ with a remainder constant.

$=(ax-2)+k/(x-1)$

Now, the last term is infinite in the limit unless $k=0$ ,

therefore $k=0$

So we are finding the limit of $(ax-2)(x-1)/(x-1)$ ,

= the limit of $ax-2$ , which we are told is $1$ , when $x=1$

Thus $a=3$ and from $(ax-2)(x-1)=ax^2+bx+2$ , we can see $b=-5$

Answer is $\boxed{-2}$