## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0^+}(\tan x)^x=1.$$
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 $$L=e^l=e^0=1.$$