jee-main 2025 Q25

jee-main · India · session1_28jan_shift2 Laws of Logarithms Simplify or Evaluate a Logarithmic Expression
Let $f ( x ) = \lim _ { \mathrm { n } \rightarrow \infty } \sum _ { \mathrm { r } = 0 } ^ { \mathrm { n } } \left( \frac { \tan \left( x / 2 ^ { r + 1 } \right) + \tan ^ { 3 } \left( x / 2 ^ { r + 1 } \right) } { 1 - \tan ^ { 2 } \left( x / 2 ^ { r + 1 } \right) } \right)$. Then $\lim _ { x \rightarrow 0 } \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { f ( x ) } } { ( x - f ( x ) ) }$ is equal to
Let $f ( x ) = \lim _ { \mathrm { n } \rightarrow \infty } \sum _ { \mathrm { r } = 0 } ^ { \mathrm { n } } \left( \frac { \tan \left( x / 2 ^ { r + 1 } \right) + \tan ^ { 3 } \left( x / 2 ^ { r + 1 } \right) } { 1 - \tan ^ { 2 } \left( x / 2 ^ { r + 1 } \right) } \right)$. Then $\lim _ { x \rightarrow 0 } \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { f ( x ) } } { ( x - f ( x ) ) }$ is equal to