Question

Нужно найти предел, не используя правило Лопиталя.

Answer 1

Ответ:

$\lim_{x \to \pi } \frac{e^{sin(x)} - e^{sin(2x)}}{\sqrt{\pi x}-\pi } = -6$

Пошаговое объяснение:

$\lim_{x \to \pi } \frac{e^{sin(x)} - e^{sin(2x)}}{\sqrt{\pi x}-\pi } = \lim_{x \to \pi } \frac{e^{sin(x)} -1- (e^{sin(2x)}-1)}{\sqrt{\pi x}-\pi } = \lim_{x \to \pi } \frac{sin(x) - sin(2x)}{\sqrt{\pi x}-\pi } = \\= \lim_{x \to \pi } \frac{sin(x)(1- 2cos(x))}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) } = \lim_{x \to \pi } \frac{sin(x-\pi )(-1- 2cos(x-\pi ))}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) } = \\=\lim_{x \to \pi } \frac{sin(x-\pi )(-3)}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) }$

$\lim_{x \to \pi } \frac{sin(x-\pi )(-3)}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) } = \lim_{x \to \pi } \frac{-3(x-\pi )}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) } = \lim_{x \to \pi } \frac{-3(\sqrt{x} -\sqrt{\pi } )(\sqrt{x} +\sqrt{\pi } )}{\sqrt{\pi }(\sqrt{x} -\sqrt{\pi } ) } = \\ =\lim_{x \to \pi } \frac{-3(\sqrt{x} +\sqrt{\pi } )}{\sqrt{\pi } } = \frac{-3*2\sqrt{\pi } }{\sqrt{\pi } } = -6$