$\lim\limits_{x \rightarrow 2} \frac{\tan \left(e^{x-2}-1\right)}{\ln (x-1)}$

2 -2 1 -1

1

Put x - 2 = t $\lim\limits_{t \rightarrow 0} \frac{\tan \left(e^t-1\right)}{ln(t+1)} . \frac{\left(e^t-1\right)}{\left(e^t-1\right)} . \frac{t}{t}=1$ Hence (3) is the correct answer.