Đáp án:
f) $\lim\limits_{x\to 0^+}(\sin x)^x = 1$
g) $\lim\limits_{x\to 0^+}(\cot x)^x=1$
Giải thích các bước giải:
$\begin{array}{l}f)\quad \lim\limits_{x\to 0^+}(\sin x)^x\\ = \lim\limits_{x\to 0^+}e^{\displaystyle{\ln(\sin x)^x}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\ln(\sin x)^x}}\\ = e^{\displaystyle{(\lim\limits_{x\to 0^+}x\ln\sin x)}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln\sin x}{\tfrac1x}}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\dfrac{\cos x}{\sin x}}{-\tfrac{1}{x^2}}}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}-\dfrac{x^2\cos x}{\sin x}}}\\ = e^{\displaystyle{-\lim\limits_{x\to 0^+}\cos x\cdot \lim\limits_{x\to 0^+}\dfrac{x^2}{\sin x}}}\\ = e^{\displaystyle{-\cos0\cdot\lim\limits_{x\to 0^+}\dfrac{2x}{\cos x}}}\\ = e^{\displaystyle{-1\cdot\dfrac{2.0}{\cos0}}}\\ = e^0 = 1\\ g)\quad \lim\limits_{x\to 0^+}(\cot x)^x\\ = = \lim\limits_{x\to 0^+}e^{\displaystyle{\ln(\cot x)^x}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\ln(\cot x)^x}}\\ = e^{\displaystyle{(\lim\limits_{x\to 0^+}x\ln\cot x)}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln\cot x}{\tfrac1x}}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{-\dfrac{1}{\sin^2x\cot x}}{-\tfrac{1}{x^2}}}}\\ = e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{x^2}{\sin^2x\cot x}}}\\ = e^{\displaystyle{\dfrac{\lim\limits_{x\to 0^+}\left(\dfrac{x}{\sin x}\right)^2}{\lim\limits_{x\to 0^+}\cot x}}}\\ = e^{\displaystyle{\dfrac{\left(\lim\limits_{x\to 0^+}\dfrac{x}{\sin x}\right)^2}{\lim\limits_{x\to 0^+}\cot x}}}\\ = e^{\displaystyle{\dfrac{\left(\lim\limits_{x\to 0^+}\dfrac{1}{\cos x}\right)^2}{\lim\limits_{x\to 0^+}\cot x}}}\\ = e^{\displaystyle{\dfrac{\left(\dfrac{1}{\cos0}\right)^2}{\cot0}}}\\ = e^{\displaystyle{\tfrac{1}{+\infty}}}\\ = e^0 = 1 \end{array}$
