_ x 0 x-x+ x^ 3 6 x^ 5 =
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
\[\lim _{x \rightarrow 0}\left\{\frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}}\right\}=\]
- 1/120
- -1/120
- 1/20
- None of these
Answer
(A) 1/120
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