Danh sách câu hỏi
[mathop lim limitsx to 0 dsin axbx - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin ax}}{{bx}}\)
[mathop lim limitsx to 0 căn 2x + 1 - căn [3]x^2 + 1sin x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - \sqrt[3]{{{x^2} + 1}}}}{{\sin x}}\)
[mathop lim limitsx to 0 dsin axbx - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin ax}}{{bx}}\)
[mathop lim limitsx to 0 căn 1 + tan x - căn 1 + sin x x^3 - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \tan x} - \sqrt {1 + \sin x} }}{{{x^3}}}\)
[mathop lim limitsx to 0 sin 5x.sin 3x.sin x45x^3 - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x.\sin 3x.\sin x}}{{45{x^3}}}\)
[mathop lim limitsx to 0 1 - căn [3]cos xtan ^2x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
[mathop lim limitsx to 0 sin x.sin 2x...sin nxn!x^n( n in N^* ) - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x.\sin 2x...\sin nx}}{{n!{x^n}}}\,\,\,\left( {n \in {\mathbb{N}^*}} \right)\)
[mathop lim limitsx to 0 1 - cos 4xx.sin 2x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 4x}}{{x.\sin 2x}}\)
[mathop lim limitsx to 0 1 - cos ^3xx.sin x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}x}}{{x.\sin x}}\)
[mathop lim limitsx to 0 cos ax - cos bxx^2 - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}}\)
[mathop lim limitsx to 0 sin 6x + sin 4xsin 3x + sin 5x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 6x + \sin 4x}}{{\sin 3x + \sin 5x}}\)
[mathop lim limitsx to 0 1 - sin 2x - cos 2x1 - sin 2x + cos 2x - Tự Học 365] \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sin 2x - \cos 2x}}{{1 - \sin 2x + \cos 2x}}\)
[mathop lim limitsx to a sin ^2x - sin ^2ax^2 - a^2 - Tự Học 365] \(\mathop {\lim }\limits_{x \to a} \frac{{{{\sin }^2}x - {{\sin }^2}a}}{{{x^2} - {a^2}}}\)
[mathop lim limitsx to 3^ + d căn x^2 - 4x + 3 3 - x - Tự Học 365] \(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\sqrt {{x^2} - 4x + 3} }}{{3 - x}}\)
[mathop lim limitsx to 1^ - d căn 1 - x + x - 1 căn x^2 - x^3 - Tự Học 365] \(\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{\sqrt {1 - x} + x - 1}}{{\sqrt {{x^2} - {x^3}} }}\)
