How do you evaluate continuity for a g(x) function with given conditions?

Question

Given $g(x)$ such that

$ \frac{af(x)}{(x^2-5x+6)}+4x$ if $x<2$ and

2 if $ x=2$

$\frac{bf(x)}{sin(x-2)}+a$ if $ x>2$

Find the values for continuity with the given conditions:

Condition 1: $$\lim_{x\rightarrow 2}\frac{f(x)}{x-2}=6$$

And condition 2:

$$f(x)=0 \Leftrightarrow x=2$$

Here is what I've tried by myself: Left side: $$\lim_{x \rightarrow 2^-} \frac{af(x)}{(x^2-5x+6)}+4x=$$ $$\lim_{x \rightarrow 2^-} \frac{a}{(x-3)} \lim_{x \rightarrow 2^-} \frac{f(x)}{(x-2)}+\lim_{x\rightarrow 2^-}4x=$$

$$\lim_{x \rightarrow 2^-} \frac{a6}{(x-3)} +4x= $$ $$-6a+8$$ How do you consider the other condition for the right side to evaluate continuity? I think that I could do it like the following : $$\lim_{x \rightarrow 2^+} f(x)= 0$$
but I'm not sure cause it change the whole limit from the right side directly to 0. What do you think?. Suggestions will be welcome. Thank you.

Answer 1

$ g$ is continuous at $ x=2$ if $$\lim_{2-}g=\lim_{2^+}g=g(2)=2$$

or

$$\lim_{2^-}\frac{af(x)}{(x-2)(x-3)}+8=2$$

and $$\lim_{2^+}\frac{f(x)(x-2)b}{(x-2)\sin(x-2)}+a=2$$

we get $$-6a+8=2$$ and $$6b+a=2$$ thus $$\boxed{a=1\;\;;\;\;b=\frac 16}$$