Question

Вычислить пределы:
$1.\ \lim_{x \to 1} \dfrac{3^{5x-3}-3^{2x^2}}{tg\ \pi x} \\\\2.\ \lim_{x \to 0}\dfrac{1+sin\ x\cdot cos\ ax}{1+sin\ x\cdot cos\ bx}$

Answer 1

$\displaystyle \lim_{x \to1}\dfrac{3^{5x-3}-3^{2x^2}}{{\rm tg}\pi x}=\lim_{x \to1}\dfrac{3^{2x^2}\Big(3^{5x-3-2x^2}-1\Big)\cdot \Big(5x-3-2x^2\Big)}{{\rm tg}\pi x\cdot \Big(5x-3-2x^2\Big)}=\\ \\ \\ =9\ln 3\lim_{x \to1}\dfrac{5x-3-2x^2}{{\rm tg}\pi x}=9\ln3\lim_{x \to1}\frac{5x-3-2x^2}{\sin \pi x}\cdot \cos \pi x=\\ \\ \\ =9\ln 3\lim_{x \to1}\frac{5x-3-2x^2}{\sin \pi x}\cdot (-1)=9\ln 3\lim_{x \to1}\frac{2x^2-5x+3}{\sin \pi x}=$

$\displaystyle =9\ln 3\lim_{x \to1}\frac{(x-1)(2x-3)}{\sin(\pi +\pi(x-1))}=9\ln 3\lim_{x \to1}\frac{(x-1)(2x-3)}{-\sin \pi(x-1)}=\\ \\ \\ =9\ln 3\lim_{x \to1}\frac{(x-1)(2\cdot 1-3)}{-\pi(x-1)}=\frac{9\ln 3}{\pi}$

$2.\displaystyle~\lim_{x \to0}\dfrac{1+\sin x\cos ax}{1+\sin x\cos bx}=\dfrac{1+0}{1+0}=1$