## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to1^+}\ln x\tan(\pi x/2)=-\frac{2}{\pi}$$
$$A=\lim_{x\to1^+}\ln x\tan(\pi x/2)$$ $$A=\lim_{x\to1^+}\frac{\ln x\sin(\pi x/2)}{\cos(\pi x/2)}$$ Because $\lim_{x\to1^+}[\ln x\sin(\pi x/2)]=\ln 1\sin(\frac{\pi\times1}{2})=0\times\sin(\frac{\pi}{2})=0\times1=0$ and $\lim_{x\to1^+}\cos(\pi x/2)=\cos(\frac{\pi\times1}{2})=\cos\frac{\pi}{2}=0$, this form is an indeterminate one of $\frac{0}{0}$. We are hence eligible to use L'Hospital's Rule: $$A=\lim_{x\to1^+}\frac{\frac{\sin(\pi x/2)}{x}+\ln x\cos(\pi x/2)(\pi x/2)'}{-\sin(\pi x/2)(\pi x/2)'}$$ $$A=\lim_{x\to1^+}\frac{\sin(\pi x/2)/x+(\pi/2)\ln x\cos(\pi x/2)}{-(\pi/2)\sin(\pi x/2)}$$ $$A=\frac{\frac{\sin(\frac{\pi\times1}{2})}{1}+(\pi/2)\ln1\cos(\frac{\pi\times1}{2})}{-(\pi/2)\sin(\frac{\pi\times1}{2})}$$ $$A=\frac{\sin(\pi/2)+(\pi/2)\times0\times\cos(\pi/2)}{-(\pi/2)\sin(\pi/2)}$$ $$A=\frac{1+0}{-(\pi/2)\times1}$$ $$A=-\frac{1}{\frac{\pi}{2}}$$ $$A=-\frac{2}{\pi}$$