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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

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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 $b$.

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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.

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