JEE Mains · Maths · STD 11 - 12. limits
Let \(p = \mathop {\lim }\limits_{x \to 0 + } {\left( {1 + {{\tan }^2}\sqrt x } \right)^{\frac{1}{{2x}}}},\) then \(\log p = \) . . .
- A \(\frac{1}{2}\;\;\)
- B \(\frac{1}{4}\)
- C \(2\)
- D \(1\)
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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