Câu hỏi:
Tính giới hạn sau \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} – 2x}}{{x – 1}}\).
-
A.
\( – \frac{1}{2}.\) -
B.
\(2.\) -
C.
\(3.\) -
D.
\( – \frac{3}{2}.\)
Lời giải tham khảo:
Đáp án đúng: D
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} – 2x}}{{x – 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x + 2} – 2x} \right)\left( {\sqrt {2x + 2} + 2x} \right)}}{{\left( {x – 1} \right)\left( {\sqrt {2x + 2} + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{2x + 2 – 4{x^2}}}{{\left( {x – 1} \right)\left( {\sqrt {2x + 2} + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{ – 2\left( {x – 1} \right)\left( {2x + 1} \right)}}{{\left( {x – 1} \right)\left( {\sqrt {2x + 2} + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{ – 2\left( {2x + 1} \right)}}{{\sqrt {2x + 2} + 2x}}\\ = \frac{{ – 2.\left( {2.1 + 1} \right)}}{{\sqrt {2.1 + 2} + 2.1}} = – \frac{3}{2}\end{array}\)
Chọn D.
Trả lời