Bài giải của hocmai.toanhoc ( Trịnh Hào Quang)
Bài 1:
[TEX]{\lim }\limits_{x \to 2} \frac{{x^2 - 4}}{{c{\rm{os}}\frac{\pi }{4}x}}[/TEX]
Giải: Đặt t=x-2 => x=t+2 và (x+2)(x-2)=t(t+4)
[TEX]{\lim }\limits_{t \to 0} \frac{{t(t + 4)}}{{c{\rm{os}}\left( {\frac{\pi }{4}t + \frac{\pi }{2}} \right)}} = - \lim \limits_{t \to 0} \frac{{t(t + 4)}}{{\sin \frac{\pi }{4}t}} = - \lim \limits_{t \to 0} \frac{{t(t + 4)}}{{\frac{\pi }{4}t.\frac{{\sin \frac{\pi }{4}t}}{{\frac{\pi }{4}t}}}} = - \lim \limits_{t \to 0} \frac{{(t + 4)}}{{\frac{\pi }{4}}} = - \frac{{16}}{\pi }[/TEX]
Bài 2:
[TEX]{\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin (x - \frac{\pi }{6})}}{{\frac{{\sqrt 3 }}{2} - c{\rm{osx}}}}[/TEX]
Giải:Cũng đổi biến ta có:
[TEX]{\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin (x - \frac{\pi }{6})}}{{\frac{{\sqrt 3 }}{2} - c{\rm{osx}}}} = {\lim }\limits_{t \to 0} \frac{{\sin t}}{{\frac{{\sqrt 3 }}{2} - c{\rm{os(t + }}\frac{\pi }{6})}} = {\lim }\limits_{t \to 0} \frac{{2\sin \frac{t}{2}.c{\rm{os}}\frac{t}{2}}}{{2\sin \frac{t}{2}\left( {\sin \frac{t}{2} + c{\rm{os}}\frac{t}{2}} \right)}} = {\lim }\limits_{t \to 0} \frac{{c{\rm{os}}\frac{t}{2}}}{{\left( {\sin \frac{t}{2} + c{\rm{os}}\frac{t}{2}} \right)}} = 1[/TEX]
Bải 3:
[TEX]{\lim }\limits_{x \to 0} \left[ {\frac{2}{{\sin 2x}} - \cot x} \right][/TEX]
Giải:
[TEX]{\lim }\limits_{x \to 0} \left[ {\frac{2}{{\sin 2x}} - \cot x} \right] = {\lim }\limits_{x \to 0} {{\rm t}\nolimits} {\rm{anx}} = 0[/TEX]
Bài 4:
[TEX]{\lim }\limits_{x \to \frac{\pi }{4}} \left( {\pi - 4x} \right)\tan 2x[/TEX]
Giải:
[TEX]{\lim }\limits_{x \to \frac{\pi }{4}} \left( {\pi - 4x} \right)\tan 2x = 4{\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\left( {\frac{\pi }{4} - x} \right)}}{{\cot 2x}} = 4{\lim }\limits_{t \to 0} \frac{t}{{\cot \left( {2t + \frac{\pi }{2}} \right)}} = - 4{\lim }\limits_{t \to 0} \frac{t}{{\tan 2t}} = - 4{\lim }\limits_{t \to 0} \frac{2}{{\frac{{2t}}{{\tan 2t}}}} = - 8[/TEX]
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