Đáp án:
\(\begin{array}{l}
38)\quad \lim\limits_{x\to 0^+}\dfrac{1 - \sqrt{\cos x}}{1 - \cos^2\sqrt x}= 0\\
39)\quad \lim\limits_{x\to 0^+}\left(\cos\sqrt x\right)^{\dfrac1x}= \dfrac{1}{\sqrt e}\\
40)\quad \lim\limits_{x\to 0}\dfrac{\sqrt{2 - \sqrt{4-x^2}}}{x}= \dfrac12
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
38)\quad \lim\limits_{x\to 0^+}\dfrac{1 - \sqrt{\cos x}}{1 - \cos^2\sqrt x}\\
= \lim\limits_{x\to 0^+}\dfrac{1 - \cos x}{\left(1 + \sqrt{\cos x}\right)\left(1 +\cos\sqrt x\right)\left(1 - \cos\sqrt x\right)}\\
= \lim\limits_{x\to 0^+}\dfrac{1}{\left(1 + \sqrt{\cos x}\right)\left(1 +\cos\sqrt x\right)}\cdot\lim\limits_{x\to 0^+}\dfrac{1 - \cos x}{1 - \cos\sqrt x}\\
= \dfrac{1}{\left(1 + \sqrt{\cos 0}\right)\left(1 +\cos\sqrt 0\right)}\cdot\lim\limits_{x\to 0^+}\dfrac{\dfrac12x^2}{\dfrac12\left(\sqrt x\right)^2}\\
= \dfrac14\cdot\lim\limits_{x\to 0^+}x\\
= \dfrac14\cdot 0\\
= 0\\
39)\quad \lim\limits_{x\to 0^+}\left(\cos\sqrt x\right)^{\dfrac1x}\\
= \lim\limits_{x\to 0^+}e^{\displaystyle{\ln\left(\cos\sqrt x\right)^{\dfrac1x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln\cos\sqrt x}{x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\left(-\dfrac{\sin\sqrt x}{2\sqrt x}\right)}}\qquad \text{(l'Hôpital)}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\left(-\dfrac{\sqrt x}{2\sqrt x}\right)}}\\
= e^{\displaystyle{-\dfrac12}}\\
= \dfrac{1}{\sqrt e}\\
40)\quad \lim\limits_{x\to 0}\dfrac{\sqrt{2 - \sqrt{4-x^2}}}{x}\\
= \lim\limits_{x\to 0}\sqrt{\dfrac{2 - \sqrt{4-x^2}}{x^2}}\\
= \sqrt{\lim\limits_{x\to 0}\dfrac{2 - \sqrt{4 - x^2}}{x^2}}\\
= \sqrt{\lim\limits_{x\to 0}\dfrac{1}{2 + \sqrt{4 - x^2}}}\\
= \sqrt{\dfrac{1}{2 + \sqrt{4 - 0^2}}}\\
= \dfrac12
\end{array}\)
