A Trigo limit (1)

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0} \left[\cos \left(\frac \pi 4+x\right) \sec \left(\frac \pi 4-x\right) \right]^\frac 1x & \small \color{#3D99F6} \text{A }1^\infty \text{ case, see note below.} \\ & = \lim_{x \to 0} \exp \left[ \frac {\left(\cos \left(\frac \pi 4+x\right) \sec \left(\frac \pi 4-x\right) - 1\right)}x \right] & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \exp \left[ \frac {-\sin \left(\frac \pi 4+x\right) \sec \left(\frac \pi 4-x\right) - \cos \left(\frac \pi 4+x\right) \sec \left(\frac \pi 4-x\right) \tan \left(\frac \pi 4-x\right)}1 \right] & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = e^{-2} \end{aligned}$

$\implies \ln \left(\frac 1L \right) = \ln e^2 = \boxed{2}$

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Note: see 2nd method

For $\displaystyle \lim_{x \to a} \left[\frac {f(x)}{g(x)}\right]^{h(x)} = 1^\infty$ , $\implies \displaystyle \lim_{x \to a} \left[\frac {f(x)}{g(x)}\right]^{h(x)} = \exp \left[\lim_{x \to a} h(x) \left(\frac {f(x)}{g(x)}-1\right) \right]$