Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0^+}(\tan 2x)^x=1$$
$$A=\lim_{x\to0^+}(\tan 2x)^x$$ To take down the power of $x$ here, what we can do is to use a logarithm. However, we would take not just the logarithm, but the natural logarithm of both sides, since they are always easier to deal with when being involved with a derivative. $$\ln A=\ln[\lim_{x\to0^+}(\tan 2x)^x]$$ $$\ln A=\lim_{x\to0^+}[\ln (\tan 2x)^x]$$ $$\ln A=\lim_{x\to0^+}[x\ln(\tan 2x)]$$ $$\ln A=\lim_{x\to0^+}\Big[\frac{x\ln (\tan 2x)}{1}\Big].$$ Now we would divide both numerator and denominator by $x$, meaning that $$\ln A=\lim_{x\to0^+}\Big[\frac{\ln(\tan 2x)}{\frac{1}{x}}\Big].$$ Since $\lim_{x\to0^+}\ln(\tan 2x)=\ln(\tan (2\times0))=\ln(\tan0))=\ln0=\infty$ and $\lim_{x\to0^+}\frac{1}{x}=\infty$, we end up here with an indeterminate form of $\infty/\infty$, eligible to the application of L'Hospital's Rule: $$\ln A=\lim_{x\to0^+}\Big[\frac{\frac{1}{\tan 2x}(\tan 2x)'}{-\frac{1}{x^2}}\Big]$$ $$\ln A=-\lim_{x\to0^+}\frac{\frac{1}{\tan 2x}(2\times\sec^2 (2x))}{\frac{1}{x^2}}$$ $$\ln A=-\lim_{x\to0^+}\frac{\frac{2\sec^2(2x)}{\tan 2x}}{\frac{1}{x^2}}$$ $$\ln A=-\lim_{x\to0^+}\frac{2x^2\sec^22x}{\tan 2x}$$ $$\ln A=-\lim_{x\to0^+}\frac{2x^2\frac{1}{\cos^2 2x}}{\frac{\sin 2x}{\cos 2x}}$$ $$\ln A=-\lim_{x\to0^+}\frac{2x^2\cos 2x}{\sin 2x\cos^2 2x}$$ $$\ln A=-\lim_{x\to0^+}\frac{2x^2}{\sin 2x\cos 2x}$$ $$\ln A=-\lim_{x\to0^+}\frac{2x^2}{\frac{1}{2}\sin 4x}$$ (for $2\sin A\cos A=\sin 2A$, so $\sin A\cos A=\frac{1}{2}\sin 2A$) $$\ln A=-\lim_{x\to0^+}\frac{4x^2}{\sin 4x}.$$ For $\lim_{x\to0^+}(4x^2)=4\times0^2=0$ and $\lim_{x\to0^+}\sin 4x=\sin 0=0$, we have an indeterminate form of $\frac{0}{0}$. We use L'Hospital's Rule once more: $$\ln A=-\lim_{x\to0^+}\frac{8x}{4\cos 4x}$$ $$\ln A=-\frac{8\times0}{4\cos(4\times0)}$$ $$\ln A=-\frac{0}{4\cos0}$$ $$\ln A=-\frac{0}{4\times1}=0.$$ Consequently, $$A=e^0=1$$.