Limit _ x 2 (e^ x-2 -1 ) (x-1) =
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
\[\operatorname{Limit}_{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}=\]
- 0
- – 1
- 2
- 1
Answer
(D) 1
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