Answer
$0$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin x)}{x^2+1}$
But $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin 2x)}{x^2+1}$ does not show any indeterminate form . So, we will not apply the L'Hospital's rule .
$\lim\limits_{x \to +\infty} \dfrac{x(2+\sin x)}{x^2+1}=\lim\limits_{x \to +\infty} \dfrac{2+\sin x}{(x+\dfrac{1}{x})}$
We can see that the numerator tends to a finite value because $\sin x$ is bounded and the denominator tends to $+\infty$ when $x \to \infty $ .
Thus, we have: $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin x)}{x^2+1}=\lim\limits_{x \to +\infty} \dfrac{2+\sin x}{(x+\dfrac{1}{x})}=0$
