We're given the one-sided limit,
[tex]\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}[/tex]
Evaluating the limand directly at x = 1 gives the indeterminate from
(sin(1 - 1) - exp(1 - 1) + 1) / ln(1) = 0/0
so we can potentially solve the limit by applying L'Hopital's rule. Doing so gives
[tex]\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}=\lim_{x\to1^+}\frac{-\cos(1-x)-e^{x-1}}{\frac1x}=\frac{-\cos(0)-e^0}{\frac11}=\boxed{-2}[/tex]
