求極限 \(\lim\limits_{x \to 2} \;\sqrt {4{x^2} + 3x + 3} \) ?
詳解:直接代入 \(x = 2\),得到極限值為 \(\lim\limits_{x \to 2} \;\sqrt {4{x^2} + 3x + 3} = \sqrt {25} = 5\)