pi...

July 2022 1 36 Report

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lim (sin((x-4)/2)*tg(pi*x/8))^(x-3)
x->4

-4/pi

Answers & Comments

Alexandr130398 $\lim_{x \to 4} (sin \frac{x-4}{2} *tg \frac{ \pi x}{8} )^{x-3}= \lim_{x \to 4} ( \frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} })^{x-3} \\ \\ y=( \frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} })^{x-3} \\ \\ lny=ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} })^{x-3}=(x-3)ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} }) \\ \\ \lim_{x \to4} (lny)=$

$= \lim_{x \to 4} (x-3)ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} }) = \\ \\ =\lim_{x \to 4} (x-3)*\lim_{x \to 4} ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} }) = \\ \\ =\lim_{x \to 4} (4-3)*\lim_{x \to 4} ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} })=\lim_{x \to 4} ln(\frac{sin \frac{x-4}{2} *sin \frac{ \pi x}{8} }{cos \frac{ \pi x}{8} }) \\ \\ =$

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