תרגיל באינפי דוגמא 28.1.10

[math]\displaystyle{ \lim_{x\rightarrow \frac{\pi}{2}} 2xtg(x)-\frac{\pi}{cos(x)}= \lim_{x\rightarrow \frac{\pi}{2}} \frac{2xsin(x)}{cos(x)}-\frac{\pi}{cos(x)} = \lim_{x\rightarrow \frac{\pi}{2}} \frac{2xsin(x)-\pi}{cos(x)} = }[/math]

[math]\displaystyle{ =\lim_{\Delta x\rightarrow 0} \frac{2(\frac{\pi}{2} + \Delta x)sin(\frac{\pi}{2} + \Delta x)-\pi}{cos(\frac{\pi}{2} + \Delta x)} = \lim_{\Delta x\rightarrow 0} \frac{2(\frac{\pi}{2} + \Delta x)cos(\Delta x)-\pi}{-sin(\Delta x)} = }[/math]

[math]\displaystyle{ =\lim_{\Delta x\rightarrow 0} \frac{\pi cos(\Delta x)-\pi}{sin(\Delta x)} + 2cos(\Delta x)\frac{\Delta x}{-sin(\Delta x)}= }[/math]

[math]\displaystyle{ \lim_{\Delta x\rightarrow 0} \frac{\pi [cos(\Delta x)-1]}{\Delta x}\frac{\Delta x}{sin(\Delta x)} + 2cos(\Delta x)\frac{\Delta x}{-sin(\Delta x)}= \pi \cdot 0 \cdot 1 + 2 \cdot 1 \cdot (-1) = -2 }[/math]