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[Гост]

Моля за помощ със следната задача:
[tex]\lim_{x \to π}\frac{1-sin{\frac{x}{2}}}{cos\frac{x}{2}(cos\frac{x}{4}-sin\frac{x}{4})}[/tex]

[KOPMOPAH]

$~~\lim_{x \to π}\frac{1-\sin{\frac{x}{2}}}{\cos\frac{x}{2}(\cos\frac{x}{4}-\sin\frac{x}{4})}=\lim_{x \to π}\frac{(1-\sin{\frac{x}{2}})(\cos\frac{x}{4}+\sin\frac{x}{4})}{\cos\frac{x}{2}(\cos\frac{x}{4}-\sin\frac{x}{4})(\cos\frac{x}{4}+\sin\frac{x}{4})}=\lim_{x \to π}\frac{(1-\sin{\frac{x}{2}})(\cos\frac{x}{4}+\sin\frac{x}{4})}{\cos\frac{x}{2}(\cos^2\frac{x}{4}-\sin^2\frac{x}{4})}=$

$~~=\lim_{x \to π}\frac{(1-\sin{\frac{x}{2}})(\cos\frac{x}{4}+\sin\frac{x}{4})}{\cos^2\frac{x}{2}}=\lim_{x \to π}\frac{(1-\sin{\frac{x}{2}})(\cos\frac{x}{4}+\sin\frac{x}{4})}{1-\sin^2\frac{x}{2}}=\lim_{x \to π}\frac{\cancel{(1-\sin{\frac{x}{2}})}(\cos\frac{x}{4}+\sin\frac{x}{4})}{\cancel{(1-\sin\frac{x}{2})}(1+\sin\frac{x}{2})}=\frac{\frac{\sqrt 2}2+\frac{\sqrt 2}2}{1+1}=\cdots$

[S.B.]

Моля за помощ със следната задача:
[tex]\lim_{x \to π}\frac{1-sin{\frac{x}{2}}}{cos\frac{x}{2}(cos\frac{x}{4}-sin\frac{x}{4})}[/tex]

Още един поглед върху задачата

[tex]\lim_{x \to \pi } \frac{1 - \sin \frac{x}{2} }{\cos \frac{x}{2}(\cos \frac{x}{4} - \sin \frac{x}{4}) } =[/tex]

[tex]= \lim_{x \to \pi } \frac{ \sin^{2 } \frac{x}{4} + \cos^{2 } \frac{x}{4} - 2\sin \frac{x}{4}\cos \frac{x}{4} }{( \cos^{2 } \frac{x}{4} - \sin^{2 } \frac{x}{4})(\cos \frac{x}{4} - \sin \frac{x}{4} )} =[/tex]

[tex]= \lim_{x \to \pi } \frac{ (\cos \frac{x}{4} - \sin \frac{x}{4}) ^{2 } }{(\cos \frac{x}{4} + \sin \frac{x}{4} ) (\cos \frac{x}{4} - \sin \frac{x}{4}) ^{2 } } =[/tex]

[tex]= \lim_{x \to \pi } \frac{1}{\cos \frac{x}{4} + \sin \frac{x}{4} } = \frac{1}{\cos \frac{ \pi }{4} + \sin \frac{ \pi }{4} } = \frac{1}{ 2\frac{ \sqrt{2} }{2} } = \frac{ \sqrt{2} }{2}[/tex]