Cho hàm số

- Hướng dẫn giải

$f\left(x\right)=\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}$

Ta có

$\begin{array}{l}\underset{x\to +\infty }{\lim }f\left(x\right)=\underset{x\to +\infty }{\lim }(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4})\\ =\underset{x\to +\infty }{\lim }\frac{\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to +\infty }{\lim }\frac{\left(x^2+2x+4\right)-\left(x^2-2x+4\right)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to +\infty }{\lim }\frac{4x}{x\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+x\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}\\ =\underset{x\to +\infty }{\lim }\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=2\end{array}$

$\begin{array}{l}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\\ =\underset{x\to -\infty }{\lim }\frac{\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to -\infty }{\lim }\frac{\left(x^2+2x+4\right)-\left(x^2-2x+4\right)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{\lim }\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{\lim }\frac{4x}{-x\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}-x\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}\\ =\underset{x\to -\infty }{\lim }\frac{4}{-\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}-\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=\frac{4}{-1-1}=-2\end{array}$

Đáp án cần chọn là: D