$\lim_{n \to \infty}-x+ \sqrt[]{x^2+4040x+11}$
= $\lim_{n \to \infty}\dfrac{x^2+4040x+11-x^2}{\sqrt[]{x^2+4040x+11}+x}$
= $\lim_{n \to \infty}\dfrac{4040x+11}{x\sqrt[]{1+4040/x+11/x^2}+x}$
= $\lim_{n \to \infty}\dfrac{4040+11/x}{\sqrt[]{1+4040/x+11/x^2}+1}$
= $\dfrac{4040+0}{\sqrt[]{1+0+0}+1}$
= $2020$