Answer
The solution is
$$\lim_{x\to0^+}(\tan x)^x=1.$$
Work Step by Step
To solve this limit we will transform the expression under the limit. Since the natural exponential function and the natural logarithmic function are inverse to each other we have that $c=e^{\ln c}.$ This gives $(\tan x)^{x}=e^{\ln \left(\tan x\right)^{x}}=e^{x\ln\left(\tan x\right)}$ where we used the logarithmic rule $\ln b^c=c\ln b$. Also because the exponential function is continuous it can exchange places with the limit so we get
$$L=\lim_{x\to0^+}\left(\tan x\right)^{x}=\lim_{x\to0^+}e^{{x}\ln\left(\tan x\right)}=e^{\lim_{x\to0^+}{x}\ln\left(\tan x\right)}=e^l,$$
where we have denoted $l=\lim_{x\to0^+}{x}\ln\left(\tan x\right).$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule".
$$l=\lim_{x\to0^+}{x}\ln\left(\tan x\right)=\lim_{x\to0^+}\frac{\ln\tan x}{\frac{1}{x}}=\left[\frac{\ln\tan0^+}{\frac{1}{0^+}}\right]=\left[\frac{\ln 0^+}{\infty}\right]=\left[\frac{-\infty}{\infty}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\ln\tan x)'}{\left(\frac{1}{x}\right)'}=\lim_{x\to0^+}\frac{\frac{1}{\tan x}(\tan x)'}{-\frac{1}{x^2}}=\lim_{x\to0^+}\frac{\cot x\frac{1}{\cos^2 x}}{-\frac{1}{x^2}}=\lim_{x\to0^+}\frac{\frac{1}{\sin x\cos x}}{-\frac{1}{x^2}}=\lim_{x\to0^+}\frac{-x^2}{\sin x\cos x}=\left[\frac{-0^2}{\sin0\cos0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{(-x^2)'}{(\sin x\cos x)'}=\lim_{x\to0^+}\frac{-2x}{\sin x(\cos x)'+(\sin x)'\cos x}=\lim_{x\to0^+}\frac{-2x}{\cos^2x-\sin^2x}=\frac{-2\cdot0}{\cos^20-\sin^20}=\frac{0}{1-0}=0.$$
Putting this into the initial limit we get
$$L=e^l=e^0=1.$$
