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2. Trước hết nhắc lại không chứng minh các kết quả quen thuộc sau
\mathop {\lim }\limits_{t \to 0}\dfrac{\ln (1+t)}{t}=1
\mathop {\lim }\limits_{t \to 0}\dfrac{\sin t}{t}=1
Ta cần tính L=\mathop {\lim }\limits_{x \to 0} (\cos2x)^{\frac{1}{x\sin3x}}. Ta sẽ tính trước \ln L=\mathop {\lim }\limits_{x \to 0}\dfrac{1}{x\sin3x}\ln \cos2x. Ta có
\ln L =\mathop {\lim }\limits_{x \to 0}\dfrac{1}{x\sin3x}\ln (1+\cos2x-1)=\mathop {\lim }\limits_{x \to 0}\dfrac{\cos2x-1}{x\sin3x}.\dfrac{\ln (1+\cos2x-1)}{\cos2x-1}
Do x \to 0\Rightarrow \cos2x-1 \to 0\Rightarrow \mathop {\lim }\limits_{x \to 0}\dfrac{\ln (1+\cos2x-1)}{\cos2x-1}=1
\implies \ln L =\mathop {\lim }\limits_{x \to 0}\dfrac{\cos2x-1}{x\sin3x}=\mathop {\lim }\limits_{x \to 0}\dfrac{-2\sin^2 \dfrac{x}{2}}{x\sin3x}
\implies \ln L=-\dfrac{1}{6}.\mathop {\lim }\limits_{x \to
0}\left ( \dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}}\right )^2.\mathop {\lim }\limits_{x \to
0} \dfrac{3x}{\sin 3x}=-\dfrac{1}{6}.
Vậy \boxed{L = \dfrac{1}{\sqrt[6]{e}}}.
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