Đáp án:
$\lim\limits_{x\to 0}\dfrac{\sin x}{1 -\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)}=1$
Giải thích các bước giải:
$\quad \lim\limits_{x\to 0}\dfrac{\sin x}{1 -\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)}$
$= \lim\limits_{x\to 0}\dfrac{\sin x\left[1 +\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]}{\left[1 -\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]\left[1 +\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]}$
$=\lim\limits_{x\to 0}\dfrac{\sin x\left[1 +\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]}{1 - 2\sin^2\left(\dfrac{\pi}{4} - x\right)}$
$=\lim\limits_{x\to 0}\dfrac{\sin x\left[1 +\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]}{\cos\left(\dfrac{\pi}{2} - 2x\right)}$
$=\lim\limits_{x\to 0}\dfrac{\sin x\left[1 +\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)\right]}{\sin2x}$
$= \lim\limits_{x\to 0}\dfrac{1+\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)}{2\cos x}$
$= \dfrac{1 + \sqrt2\sin\dfrac{\pi}{4}}{2\cos0}$
$= 1$
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Cách khác:
Áp dụng quy tắc $l'Hôpital$ ta được:
$\quad \lim\limits_{x\to 0}\dfrac{\sin x}{1 -\sqrt2\sin\left(\dfrac{\pi}{4} - x\right)}$
$=\lim\limits_{x\to 0}\dfrac{\cos x}{\sqrt2\cos\left(\dfrac{\pi}{4} - x\right)}$
$=\dfrac{\cos0}{\sqrt2\cos\dfrac{\pi}{4}}$
$= 1$
