If $$f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}},$$     then $$\mathop {\lim }\limits_{x \to 2} f\left( x \right)$$   is equal to :

Solution :
$$\eqalign{ & f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}} \cr & \mathop {\lim }\limits_{x \to 2} \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}} = L \cr} $$
It is $$\frac{0}{0}$$ (undefined) condition so using L'hospital's rule
$$\eqalign{ & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\left\{ {\sin \left( {{e^{x - 2}} - 1} \right)} \right\}}}{{\left\{ {\ln \left( {x - 1} \right)} \right\}}}} \right] \cr & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \frac{{\cos \left( {{e^{x - 2}} - 1} \right).{e^{\left( {x - 2} \right)}}}}{{\frac{1}{{\left( {x - 1} \right)}}}} \cr & \Rightarrow L = \mathop {\lim }\limits_{x \to 2} \cos \left( {{e^{2 - 2}} - 1} \right){e^{2 - 2}}.\left( {2 - 1} \right) \cr & \Rightarrow L = \cos \left( 0 \right){e^0}.1 \cr & \Rightarrow L = 1 \cr} $$