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Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to 0} {{\tan 3x} \over {\tan 5x}}\) 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
\(\eqalign{& \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} – x} \right)\tan x = \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} – x} \right)\cot \left( {{\pi \over 2} – x} \right) \cr& = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left( {{\pi \over 2} – x} \right)} \over {\sin \left( {{\pi \over 2} – x} \right)}}.\cos \left( {{\pi \over 2} – x} \right) = 1 \cr} \)
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(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 \(\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{\pi \over 2} – x} \right)\tan x = \mathop {\lim }\limits_{t \to 0} t\tan \left( {{\pi \over 2} – t} \right)\)
\(= \mathop {\lim }\limits_{t \to 0} t\cot t = \mathop {\lim }\limits_{t \to 0} {t \over {\sin t}}.\cot t = 1.\)