Question

Может кто нибудь решить такое?

Answer 1

$\lim\limits _{x \to \frac{\pi}{3}}\frac{e^{sin^26x}-e^{sin^23x}}{log_3cos6x}=\lim\limits _{x \to \frac{\pi}{3}}\frac{e^{4sin^23x\cdot cos^23x}-e^{sin^23x}}{log_3(cos6x-1+1)}=\\\\\\=\lim\limits_{x \to \frac{\pi}{3}}\frac{e^{sin^23x}\cdot (e^{4cos^23x}-1)}{log_3(1+(cos6x-1))}=\lim\limits _{x \to \frac{\pi}{3}}\frac{e^{sin^23x}\cdot (e^{4cos^23x}-1)}{\frac{cos6x-1}{ln3}}=\\\\\\=\lim\limits _{x \to \frac{\pi}{3}}\frac{e^{sin^23x}\cdot (e^{4cos^23x}-1)\cdot ln3}{-2sin^23x}=\lim\limits _{x \to \frac{\pi}{3}}\frac{e^{sin^23x}\cdot (e^{4cos^23x}-1)\cdot ln3}{-2\cdot 9x^2}=-\frac{e}{2\pi ^2}(e^4-1)\cdot ln3$

$P.S.\; \; log_{b}(\alpha +1)\sim \frac{\alpha }{lnb}\; \; ,\; \; \alpha \to 0\\\\sin\alpha \sim \alpha \; \; ,\; \; \alpha \to 0$