Given limit is

\(\lim_{x\rightarrow2}\frac{\sin(x-2)}{x^2-4}\)

Solve,

\(\lim_{x\rightarrow2}\frac{\sin(x-2)}{x^2-4}=\lim_{x\rightarrow2}\frac{\sin(x-2)}{(x-2)(x+2)}\)

\(=\lim_{x\rightarrow2}(\frac{\sin(x-2)}{(x-2)})\lim_{x\rightarrow2}(\frac{1}{x+2})\)

\(=1\times(\frac{1}{4})\) \(=\frac{1}{4}\)