## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}x\sin(\pi/x)=\pi$$
$$A=\lim_{x\to\infty}x\sin(\pi/x)$$ If we leave the limit just like this, we do not have any methods to deal with it now. We also cannot do the replacement, since if we do, it would lead to $$\infty\times\sin(\frac{\pi}{\infty})$$ which has no apparent results. What we can do is to bring it back to indeterminate forms of $\frac{0}{0}$ or $\infty/\infty$. In other words, we would try to bring it to fraction form to see if it is fit for L'Hospital's Rule or not. A can also be written in this form: $$A=\lim_{x\to\infty}\frac{x\sin(\pi/x)}{1}$$ Divide both numerator and denominator by $x$: $$A=\lim_{x\to\infty}\frac{\sin(\pi/x)}{\frac{1}{x}}$$ Now we take $u=\frac{1}{x}$. As $x\to\infty$, so $u\to0$. $$A=\lim_{u\to0}\frac{\sin(\pi u)}{u}.$$ We know $\lim_{u\to0}\sin(\pi u)=\sin(0\pi)=\sin0=0$ and $\lim_{u\to0}u=0.$ Indeterminate form of $\frac{0}{0}$. We then apply L'Hospital's Rule: $$A=\lim_{u\to0}\frac{[\sin(\pi u)]'}{u'}$$ $$A=\lim_{u\to0}\frac{\pi\cos(\pi u)}{1}$$ $$A=\lim_{u\to0}(\pi\cos(\pi u))$$ $$A=\pi\cos(\pi\times0)$$ $$A=\pi\cos0$$ $$A=\pi\times1=\pi$$