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JEE Mains · Maths · STD 11 - 12. limits

ધારો કે \(p = \mathop {\lim }\limits_{x \to 0 + } {\left( {1 + {{\tan }^2}\sqrt x } \right)^{\frac{1}{{2x}}}},\) તો \(\log p = \) . . . . . થાય. . .

  1. A \(\frac{1}{2}\;\;\)
  2. B \(\frac{1}{4}\)
  3. C \(2\)
  4. D \(1\)
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Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\;\;\)

Step-by-step Solution

Detailed explanation

\({\rm{p}} = {{\rm{e}}^{\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{2}{{\left( {\frac{{{\mathop{\rm san}\nolimits} \sqrt x }}{{\sqrt x }}} \right)}^2}}} = \sqrt e \) \(\log p = \frac{1}{2}\)
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