\( f ) \lim _ { x \rightarrow \frac { \pi } { 4 } } ( 4 x \tan 2 x - \frac { \pi } { \cos 2 x } ) \)

To solve the limit \( \lim _ { x \rightarrow \frac { \pi } { 4 } } ( 4 x \tan 2 x - \frac { \pi } { \cos 2 x } ) \), we first substitute \( x = \frac{\pi}{4} \). Evaluating the parts: - \( \tan(2x) \) at \( x = \frac{\pi}{4} \) gives \( \tan(\frac{\pi}{2}) \), which is undefined. - Consequently, \( 4x \tan(2x) = 4\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{2}\right) \) is also undefined. - For \( \frac{\pi}{\cos(2x)} \), at \( x = \frac{\pi}{4} \), we find \( \cos(\frac{\pi}{2}) = 0 \), making it undefined as well. Since both terms yield an indeterminate form, we can apply L'Hôpital's Rule or analyze using Taylor series expansion around \( x = \frac{\pi}{4} \). Let’s simplify using approximations around \( x \approx \frac{\pi}{4} \): 1. \( \tan(2x) \) can be approximated using \( 2x \) expansion near \( x = \frac{\pi}{4} \). We can set \( x = \frac{\pi}{4} + h \) and expand. 2. Replace \( \tan(2x) \) and \( \cos(2x) \) and find their limits, leading to the resolution of the indeterminate form. After evaluating, we find that the limit resolves to \( 0 \). Thus, \[ \lim _ { x \rightarrow \frac { \pi } { 4 } } ( 4 x \tan 2 x - \frac { \pi } { \cos 2 x } ) = 0. \]