(10) Granice funkcji

[melon]

Oblicz granice nie korzystając z reguły de l'Hospitala

\(\lim_{x\to 1} \frac{x^n-1}{x-1}\)

\(\lim_{x\to \infty} \frac{\sin x}{x}\)

\(\lim_{x\to \frac{\pi}{2} } \ \frac{\cos x}{x-\frac{\pi}{2}}\)

\(\lim_{x\to 8} \ \frac{8-x}{\sin \frac{\pi}{8}x }\)

\(\lim_{x\to \frac{\pi}{4} } \ \frac{\cos x-\cos\frac{\pi}{4} }{\sin x-\sin \frac{\pi}{4}}\)

[radagast]

\(\lim_{x\to \frac{\pi}{2} } \ \frac{\cos x}{x-\frac{\pi}{2}}=\lim_{x\to \frac{\pi}{2} } \ \frac{\sin( \frac{ \pi }{2}- x)}{x-\frac{\pi}{2}}=-\lim_{x\to \frac{\pi}{2} } \ \frac{\sin(x- \frac{ \pi }{2})}{x-\frac{\pi}{2}}=-1\)

[radagast]

\(\lim_{x\to \infty} \frac{\sin x}{x}= \frac{ograniczona}{ \infty } =0\)

[radagast]

\(\lim_{x\to 1} \frac{x^n-1}{x-1}=\lim_{x\to 1} \frac{(x-1)(x^{n-1}+x^{n-2}+...+1)}{x-1}=n\)

[radagast]

\(\lim_{x\to 8} \ \frac{8-x}{\sin \frac{\pi}{8}x }=\lim_{x\to 8} \ \frac{8-x}{\sin ( \pi - \frac{\pi}{8}x) }=8\lim_{x\to 8} \ \frac{1- \frac{x}{8} }{\sin ( \pi - \frac{\pi}{8}x) }= \frac{8}{ \pi } \lim_{x\to 8} \ \frac{ \pi - \frac{ \pi }{8}x }{\sin ( \pi - \frac{\pi}{8}x) }=\frac{8}{ \pi }\)

[radagast]

\(\lim_{x\to \frac{\pi}{4} } \ \frac{\cos x-\cos\frac{\pi}{4} }{\sin x-\sin \frac{\pi}{4}}=\lim_{x\to \frac{\pi}{4} } \ \frac{-2\sin { \frac{4x+ \pi }{8} }sin{ \frac{4x- \pi }{8} } }{2\cos { \frac{4x+ \pi }{8} }sin{ \frac{4x- \pi }{8} } }=\lim_{x\to \frac{\pi}{4} } \ \frac{-\sin{ \frac{4x+ \pi }{8} } }{\cos { \frac{4x+ \pi }{8} } }=-\lim_{x\to \frac{\pi}{4} } \ tg{ { \frac{4x+ \pi }{8} }}=-tg \frac{ \pi }{4}=-1\)