Official solution
\[\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}\]
This is 1\[\infty \] form.
= \[{e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}\]
= \[{e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}\]
= \[{e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}\]
= \[{e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}\]
= \[{e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}\]
= e2