Answer
See graph and explanations.
Work Step by Step
See graph; when $x\to-\infty, y\to0$ or $\lim_{x\to -\infty}sin(\frac{\pi}{x^2+1})=0$;
while when $x\to\infty, y\to0$ or $\lim_{x\to \infty}sin(\frac{\pi}{x^2+1})=0$;
the asymptote of the function is $y=0$