$\displaystyle Let \ L=\lim_{x\to 0} (1+\tan x)^{\cot x}$ $\displaystyle \log \ L=\log \lim_{x\to 0} (1+\tan x)^{\cot x}$ Since logarithm is a continuous function, we can write, $\displaystyle \log L = \lim_{x\to 0} \log (1+\tan x)^{\cot x} = \lim_{x\to m}\cot x \log (1+\tan x) $ $\displaystyle \log L= \lim_{x\to 0 } \dfrac {\log (1+\tan x)}{\tan x }$ This is in the form of $\dfrac {0}{0}$, Hence we can apply L'Hopital's rule, $\displaystyle\log L=\lim_{x\to 0} \dfrac {\frac {1}{1+\tan x} \times \sec^2 x}{sec^2 x} = \lim_{x\to 0} \dfrac {1}{1+\tan x}$ Directly plugging in x=0, $\displaystyle \log L= \dfrac {1}{1+\tan 0} =1, \ L=e^1=e$ $\displaystyle \therefore \lim_{x\to 0}(1+\tan x)^{\cot x}=e$