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: 17 lis 2018, 22:20
\(\Lim_{x\to \frac{ \pi }{4} } \frac{cosx-cos \frac{ \pi }{4} }{sinx-sin \frac{ \pi }{4} }\)
\(\Lim_{x\to \frac{ \pi }{4} } \frac{\cos x-\cos \frac{ \pi }{4} }{\sin x-\sin \frac{ \pi }{4} }=^H=\Lim_{x\to \frac{ \pi }{4} } \frac{-\sin x }{\cos x}=-1\)
lub po prostu:
\(\Lim_{x\to \frac{ \pi }{4} } \frac{\cos x-\cos \frac{ \pi }{4} }{\sin x-\sin \frac{ \pi }{4} }=\Lim_{x\to \frac{ \pi }{4} } \frac{-2\sin \frac{x+ \frac{\pi}{4} }{2}\sin \frac{x- \frac{\pi}{4} }{2}}{2\sin \frac{x- \frac{\pi}{4} }{2}\cos \frac{x+\frac{\pi}{4} }{2}}=\Lim_{x\to \frac{ \pi }{4} } \frac{-\sin \frac{x+ \frac{\pi}{4} }{2}}{\cos \frac{x+\frac{\pi}{4} }{2}}=\frac{-\sin \frac{\pi}{4}} {\cos \frac{\pi}{4} }=-1\)
można jeszcze tak:
\(\Lim_{x\to \frac{ \pi }{4} } \frac{\cos x-\cos \frac{ \pi }{4} }{\sin x-\sin \frac{ \pi }{4} }=\Lim_{x\to \frac{ \pi }{4} } \frac{2\cos x- \sqrt{2} }{2\sin x-\sqrt{2}} \cdot \frac{2\sin x+\sqrt{2}}{2\sin x+\sqrt{2}} \cdot \frac{2\cos x+ \sqrt{2} }{2\cos x+ \sqrt{2} }=\Lim_{x\to \frac{ \pi }{4} } \frac{4\cos ^2x- 2 }{4\sin^2 x-2} \cdot \frac{2\sin x+\sqrt{2}}{1} \cdot \frac{1}{2\cos x+ \sqrt{2} }=\\
\Lim_{x\to \frac{ \pi }{4} } \frac{2\cos ^2x- 1 }{2\sin^2 x-1} \cdot \frac{2\sin x+\sqrt{2}}{2\cos x+ \sqrt{2} }=\Lim_{x\to \frac{ \pi }{4} } \frac{\cos 2x}{-\cos 2x} \cdot \frac{2\sin x+\sqrt{2}}{2\cos x+ \sqrt{2} }=\Lim_{x\to \frac{ \pi }{4} } \frac{1}{-1} \cdot \frac{2\sin x+\sqrt{2}}{2\cos x+ \sqrt{2} }= -\frac{ \sqrt{2}+ \sqrt{2}}{\sqrt{2}+ \sqrt{2}}=-1\)