A. − π
B. − 1
C. 0
D. 1
E. π
[ 5-7225 - op net sinds 4.1.13-(E)-20.11.2024 ]
Translation in E N G L I S H
A. − π
B. − 1
C. 0
D. 1
E. π
Oplossing - Solution
1
ste
manier : zonder de regel van de L'Hospital
2
de
manier : met de regel van de L'Hospital
}=\displaystyle\:\lim_{x\to&space;1}\:\frac{\tan\pi&space;x}{1-x}=\displaystyle\:\lim_{x\to&space;1}\:\frac{\sin\pi&space;x}{os\pi&space;x(1-x)} "\\\displaystyle\:\lim_{x\to 1}\:\frac{\tan\pi x}{1-x}=\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi x}{os\pi x(1-x)}=\displaystyle\:\lim_{x\to 1}\:\frac{\tan\pi x}{1-x}=\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi x}{os\pi x(1-x)}\\=\displaystyle\:\lim_{x\to 1}\:\frac{1}{\cos\pi x}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin(\pi-\pi x)}{1-x}=\frac{1}{\cos\pi}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi(1-x)}{1-x}\\=\frac{\pi}{\cos\pi}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi(1-x)}{\pi(1-x)}=\frac{\pi}{-1}=-\pi")
}{1-x}=\frac{1}{\cos\pi}.\displaystyle\:\lim_{x\to&space;1}\:\frac{\sin\pi(1-x)}{1-x}\\=\frac{\pi}{\cos\pi}.\displaystyle\:\lim_{x\to&space;1}\:\frac{\sin\pi(1-x)}{\pi(1-x)}=\frac{\pi}{-1}=-\pi&space; "\\\displaystyle\:\lim_{x\to 1}\:\frac{\tan\pi x}{1-x}=\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi x}{os\pi x(1-x)}=\displaystyle\:\lim_{x\to 1}\:\frac{\tan\pi x}{1-x}=\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi x}{os\pi x(1-x)}\\=\displaystyle\:\lim_{x\to 1}\:\frac{1}{\cos\pi x}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin(\pi-\pi x)}{1-x}=\frac{1}{\cos\pi}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi(1-x)}{1-x}\\=\frac{\pi}{\cos\pi}.\displaystyle\:\lim_{x\to 1}\:\frac{\sin\pi(1-x)}{\pi(1-x)}=\frac{\pi}{-1}=-\pi")
2de manier : met de regel van de L'Hospital
\overset{H}{=}\displaystyle\:\lim_{x\to&space;1}\:\frac{\frac{1}{\cos^2\pi&space;x}.\pi}{-1}=\frac{\frac{1}{(-1)^2}.\pi}{-1}=-\pi&space; "\\\displaystyle\:\lim_{x\to 1}\:\frac{\tan\pi x}{1-x}\left(=\frac00\right)\overset{H}{=}\displaystyle\:\lim_{x\to 1}\:\frac{\frac{1}{\cos^2\pi x}.\pi}{-1}=\frac{\frac{1}{(-1)^2}.\pi}{-1}=-\pi")
