Answer
$\dfrac{2}{3}$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} \dfrac{2x-\sin x}{3x+\sin x}$
But $\lim\limits_{x \to +\infty} \dfrac{2x-\sin x}{3x+\sin x}$ does not show any indeterminate form . So, we will not apply the L'Hospital's rule .
$\lim\limits_{x \to +\infty} \dfrac{2x-\sin x}{3x+\sin x}=\dfrac{2-\lim\limits_{x \to +\infty} \dfrac{\sin x}{x}}{3+\lim\limits_{x \to +\infty} \dfrac{\sin x}{x}}$
or, $=\dfrac{2-0}{3+0}$
Thus, we have $\lim\limits_{x \to +\infty} \dfrac{2x-\sin x}{3x+\sin x}=\dfrac{2}{3}$
