Question

Sa se calculeze limita ​

Answer 1

$\lim\limits_{x\to\infty} \dfrac{x\sin\frac{1}{x}}{x^2+1} = \lim\limits_{x\to\infty}\left( \dfrac{\sin\frac{1}{x}}{\frac{1}{x}}\cdot \dfrac{1}{x^2+1}\right) = \\ \\ = \lim\limits_{x\to\infty}\dfrac{\sin\frac{1}{x}}{\frac{1}{x}} \cdot \lim\limits_{x\to\infty}\dfrac{1}{x^2+1} = 1\cdot 0 = \boxed{0}$