Use the limit theorem and the properties of limits to find the limit.

Answer:AStep-by-step explanation:First, simplify the expression [tex]\dfrac{x}{x^2+1}+\dfrac{2x^2}{x^3+x}=\dfrac{x}{x^2+1}+\dfrac{2x^2}{x(x^2+1)}=\dfrac{x^2+2x^2}{x(x^2+1)}=\dfrac{3x^2}{x(x^2+1)}.[/tex]Then the limit is[tex]\lim_{x \to \infty} \dfrac{3x^2}{x(x^2+1)}.[/tex]Since the denominator of this fraction has greater power (the power of the denominator is 3) than numerator has (the power of numerator is 2), by the limit theorem this limit is equal to 0.