Let, $$ L=\lim_{x\to 0} (1+tanx)^{cotx} $$

Taking log on both sides,

$$ \therefore logL \;= \; \lim_{x\to 0} {cotx} log(1+tanx) $$

$$ = \lim_{x\to 0} \dfrac{ log(1+tanx)}{tan x} \\ \; \\ \; \\ = 1 \; \; \; \ \ \ \ \ \ \ \; \{ \because \lim_{x\to 0} \dfrac{ log(1+X)}{X} \;=\; 1 \} $$

$$ \therefore L \;=\; e^1 \;=\; e $$

$$ \lim_{x\to 0} (1+tanx)^{cotx} \;=\; e $$