How to solve this without L'Hopital?

Question

$$f(x)=\frac{\sin(x-2)}{x^2+x-6}$$ is not defined at $x=2$ and $x=-3$, how can I solve this without using de l'Hôpital's theorem to be defined continuous at $x=2$?

Answer 1

Since $\sin x\sim x$ then $\sin (x-2) \sim x-2$, which implies that $$\lim_{x\to 2}\frac{\sin (x-2)}{(x-2) (x+3)}=\lim_{x\to 2}\frac{1}{x+3}=\frac{1}{5}, $$so if you define $f(2) =1/5$ then its continuous in that point. For $x=-3$ you cannot extens continuously the function since the limit is $-\infty$.