詳解：這極限直觀為 \(\infty  - \infty \)，極限值無法直接判斷，須將 \(\sqrt {{x^2} + 6x}  - x\) 有理化

\(\lim\limits_{x \to \infty } (\sqrt {{x^2} + 6x}  - x) =\lim\limits_{x \to \infty } \frac{{\sqrt {{x^2} + 6x}  - x}}{1}\)  \(\lim\limits_{x \to \infty } \frac{{\left( {\sqrt {{x^2} + 6x}  - x} \right)\left( {\sqrt {{x^2} + 6x}  + x} \right)}}{{\left( {\sqrt {{x^2} + 6x}  + x} \right)}}\)

\(\lim\limits_{x \to \infty } \frac{{{x^2} + 6x - {x^2}}}{{\sqrt {{x^2} + 6x}  + x}}\) \(\lim\limits_{x \to \infty } \frac{{6x}}{{\sqrt {{x^2} + 6x}  + x}}\)\( = 3\)

說明：\(\lim\limits_{x \to \infty } \frac{{6x}}{{\sqrt {{x^2} + 6x}  + x}} = \lim\limits_{x \to \infty } \frac{6}{{\frac{{\sqrt {{x^2} + 6x} }}{x} + 1}}\lim\limits_{x \to \infty } \frac{6}{{\sqrt {1 + \frac{6}{x}}  + 1}} = 3\)