Calculate $a$ and $b$ in
$$\lim_{x \to 1} \frac{ax^2+(3a+1)x+3}{bx^2+(2-b)x-2} = \frac{3}{2}$$
I tried this
$$\lim_{x \to 1} \frac{(ax+1)(x+3)}{(bx+2)(x-1)} = \frac{3}{2}$$
but I could not see the next step
I tried to look but it did not help. Solve for $a$ and $b$ in a limit and Find A and B in this limit
Well, having a singularity in $1$, the only vale of $a$ that makes this limit exist is $a=-1$ (canceling out the denominator).
With that value, limit exists and equals $$-\frac{1+3}{b+2}.$$
You can now equate that to $3/2$ and find the value of $b$.
Your first step
$$\lim_{x \to 1} \frac{ax^2+(3a+1)x+3}{bx^2+(2-b)x-2} =\lim_{x \to 1} \frac{(ax+1)(x+3)}{(bx+2)(x-1)}$$
is correct, now observe that you need to cancel out the term $(x-1)$ in order to have a finite limit.
