$$\lim_{x\to 0^{+}} (\tan x)^x$$
$$\lim_{x\to 0^{+}} e^{\ln((\tan x)^x)}=\lim_{x\to 0^{+}} e^{x\ln(\tan x)}=\lim_{x\to 0^{+}} e^{x[\ln(\sin x)-\ln(\cos x)]}$$
We can continue to create an expression that may help us use L'Hospital but it does not seem to be correct
P.S can we write:
$$1=(\frac{-1}{-1})^x\leq \lim_{x\to 0^+}\Bigl(\frac{\sin x}{\cos x}\Bigr)^x\leq \Bigl(\frac{1}{1}\Bigr)^x=1$$?
Best Answer
$\lim_{x \to 0+} x \ln (\tan x)=\lim_{x \to 0+} \frac {\ln (\tan x)} {1/x}=-\lim_{x \to 0+} \frac {\sec^{2}x} {\tan x /x^{2}}$. You can write this as $-\lim_{x \to 0+} \frac {x^{2}} { \sin x \cos x}$ and this limit is $0$. (Why?). Hence the given limit is $e^{0}=1$.