mathop lim limitsx to 0 cos ax - cos bxx^2
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}}\)
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - 1 + 1 - \cos bx}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - 1}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\cos bx - 1}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{{\sin }^2}\frac{{ax}}{2}}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{{bx}}{2}}}{{{x^2}}}\\ = - 2\mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\sin \frac{{ax}}{2}}}{{\frac{{ax}}{2}}}} \right]^2}.\frac{{{a^2}}}{4} + 2\mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\sin \frac{{bx}}{2}}}{{\frac{{bx}}{2}}}} \right]^2}.\frac{{{b^2}}}{4}\\ = - \frac{{{a^2}}}{2} + \frac{{{b^2}}}{2} = \frac{{{b^2} - {a^2}}}{2}\end{array}\)
Chọn D.
