Question

Вычислить пределы:

Answer 1

$\displaystyle \lim_{x \to \infty}\left(\dfrac{1+2x}{3+2x}\right)^{-x}=\lim_{x \to \infty}\left(1-\dfrac{2}{3+2x}\right)^{(-x)\cdot \left(-\frac{2}{3+2x}\right)\cdot \left(-\frac{3+2x}{2}\right)}=\\ \\ \\ =e^{\lim_{x \to \infty}(-x)\cdot (-\frac{2}{3+2x})}=e^{\lim_{x \to \infty}\frac{2x}{3+2x}}=e$

$\displaystyle \lim_{x \to 0}\left(\frac{1+x\cdot 3^x}{1+x\cdot 7^x}\right)^{1/{\rm tg}^2x}=\lim_{x \to 0}\left(1+\frac{1+x\cdot 3^x}{1+x\cdot 7^x}-1\right)^{1/{\rm tg}^2x}=\\ \\ \\ =\lim_{x \to 0}\left(1+\frac{x\cdot 3^x-x\cdot 7^x}{1+x\cdot 7^x}\right)^\big{\frac{1}{{\rm tg}^2x}\cdot \frac{x\cdot 3^x-x\cdot 7^x}{1+x\cdot 7^x}\cdot \frac{1+x\cdot 7^x}{x\cdot 3^x-x\cdot 7^x}}=\\ \\ \\ =e^\big{\lim_{x \to 0}\frac{x\cdot 3^x-x\cdot 7^x}{(1+x\cdot 7^x)x^2}}=e^\big{\lim_{x \to 0}\frac{3^x-7^x}{x}}=$

$=\displaystyle e^\big{\lim_{x \to 0}\frac{7^x\left((\frac{3}{7})^x-1\right)}{x}}=e^\big{\ln\frac{3}{7}}=\dfrac{3}{7}$