Answer
$\pi$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to + \infty} x \sin (\dfrac{\pi}{x})=\lim\limits_{x \to + \infty} \dfrac{\sin (\pi/x)}{1/x}$.
But $\lim\limits_{x \to + \infty} \dfrac{\sin (\pi/x)}{1/x}=\dfrac{0}{0}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to + \infty} \dfrac{\sin (\pi/x)}{1/x}=\lim\limits_{x \to + \infty} \dfrac{\cos (\pi/x)(-\pi/x^2)}{(-1/x^2)} \\ =\lim\limits_{x \to + \infty} \pi (\cos \dfrac{\pi}{x}) \\=\pi \times \cos (0)\\=\pi$
