[tex]\lim_{x\rightarrow \frac{\prod }{2}}(\frac{1}{cosx}-tanx)[/tex]
@Tiến Phùng đây anh ơi
$\displaystyle \lim_{x \to \frac{\pi}{2}} (\frac{1}{cosx}-tanx)= \displaystyle \lim_{x \to \frac{\pi}{2}} (\frac{1}{cosx}-\frac{sinx}{cosx})= \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-sinx}{cosx} = \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-2sin\frac{x}{2}.cos\frac{x}{2}}{cos(2.\frac{x}{2})}= \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{sin^{2}\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}+cos^{2}\frac{x}{2}}{cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}$
$= \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{(sin\frac{x}{2}-cos\frac{x}{2})^{2}}{(cos\frac{x}{2}-sin\frac{x}{2})(cos\frac{x}{2}+sin\frac{x}{2})}= \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{sin\frac{x}{2}-cos\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}= \frac{sin\frac{\pi}{4}-cos\frac{\pi}{4}}{cos\frac{\pi}{4}+sin\frac{\pi}{4}}=\frac{0}{\sqrt{2}}=0$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
