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**Haven** But the function is of indeterminate form. I.e, $\displaystyle 0*\infty $ so we can rewrite the limit so it is of the form $\displaystyle \frac{0}{0} $

$\displaystyle x\sin{\frac{\pi}{x}} = \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}}$

we see $\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0} $

apply L'Hospital's Rule

$\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =

\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}$

Since $\displaystyle \lim_{x\rightarrow\infty} \frac{1}{x} = 0 $

$\displaystyle \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$