## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to 2\pi}\frac{x\sin x+x^2-4\pi^2}{x-2\pi}=6\pi.$$
We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to 2\pi}\frac{x\sin x+x^2-4\pi^2}{x-2\pi}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 2\pi}\frac{(x\sin x+x^2-4\pi^2)'}{(x-2\pi)'}=\lim_{x\to 2\pi}\frac{(x)'\sin x+x(\sin x)' +2x}{1}=\lim_{x\to 2\pi}(\sin x+x\cos x+2x)=\sin 2\pi+2\pi\cos2\pi+2\cdot2\pi=0+2\pi\cdot1+4\pi=6\pi.$$