Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)
\(=-1\)
lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)
Thay x vào là xong
