$$f(x) = \frac{(x-3)\cdot\ln(x-2)}{1-\cos(3-x)}$$
Is there any procedure to solving this expression? The problem right now is the fact that $\lim_{x\rightarrow3}f(x) = [\frac{0}{0}]$
I've tried multiplying by the conjuage of the denominator so I had $$\lim_{x\rightarrow3} = \frac{(x-3)\cdot\ln(x-2)(1+\cos(3-x))}{\sin^2(3-x)}$$ But of course, that didn't do much. Wolfram tells me $\lim_{x\rightarrow3}\frac{x-3}{\sin(3-x)} = -1$, however, I'd then have to extend by another $(x-3)$ and when I tried, it didn't make it any easier.
Answer as community wiki, according to André Nicolas tips:
Let $t = x-3$:
$$\lim_{x\rightarrow3}f = \lim_{t\rightarrow0} \frac{t\cdot\ln(t+1)}{1-\cos t} =\lim_{t\rightarrow 0} \frac{t\cdot\ln(t+1)\cdot(1+\cos t)}{\sin^2t} = \lim_{t\rightarrow0} \frac{\ln(t+1)}{t}\cdot\frac{t}{\sin t}\cdot 2 = 2$$