Tìm các giới hạn sau
a) lim b) \mathop {\lim }\limits_{x \to 0} {{\cos 2x - 1} \over {{{\sin }^2}3x}}
c) \mathop {\lim }\limits_{x \to 0} {{\tan x - \sin x} \over {{x^3}}} d) \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} - x} \right)\tan x
a) {3 \over 5}; b) - {2 \over 9}; c) {1 \over 2};
\bullet Cách 1
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\eqalign{& \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} - x} \right)\tan x = \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} - x} \right)\cot \left( {{\pi \over 2} - x} \right) \cr& = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left( {{\pi \over 2} - x} \right)} \over {\sin \left( {{\pi \over 2} - x} \right)}}.\cos \left( {{\pi \over 2} - x} \right) = 1 \cr}
(Vì \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{\pi \over 2} - x} \over {\sin \left( {{\pi \over 2} - x} \right)}} = 1 và \mathop {\lim }\limits_{x \to {\pi \over 2}} \cos \left( {{\pi \over 2} - x} \right) = \cos 0 = 1 )
\bullet Cách 2. Đặt {\pi \over 2} - x = t thì khi x \to {\pi \over 2} ta sẽ có t \to 0.
Vậy \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} - x} \right)\tan x = \mathop {\lim }\limits_{t \to 0} t\tan \left( {{\pi \over 2} - t} \right)
= \mathop {\lim }\limits_{t \to 0} t\cot t = \mathop {\lim }\limits_{t \to 0} {t \over {\sin t}}.\cot t = 1.