Define f(x)=lim n→∞((x+1) x^2n+x^2/(x^2n+1) for every real x for which the limit exists.

For $|x|<1$, $x^{2n}\to0$ so $f(x)=x^2$; for $|x|>1$, $x^{2n}\to\infty$ so $f(x)=x+1$; at $x=1$ the expression is $\frac{2+1}{2}=\frac32$, and at $x=-1$ it is $\frac{0+1}{2}=\frac12$. At $x=1$ the one-sided limits are $1$ and $2$, and at $x=-1$ they are $1$ (from the right) and $0$ (from the left), and neither matches the function value, so there are exactly two discontinuities and (A) is true. Splitting the integral, $\int_{-2}^{-1}(x+1)dx+\int_{-1}^{1}x^2dx+\int_1^2(x+1)dx=-\frac12+\frac23+\frac52=\frac83\ne3$, so (B) is false. $f(1)+f(-1)=\frac32+\frac12=2$, so (C) is false. Near $0$, $f(x)=x^2$, so $f'(0)=0$, not $1$, and (D) is false. A high-power numerical evaluation agrees.