Задача 55824 ...

5f92738380c7303e80e65d67

26.11.2020 13:44:49

lim (e^(tgx) - 1)/(tgx - x)
x→0

5f3ea7e3faf909182968ddd9

26.11.2020 20:23:23

★

[m]lim_{x → 0}\frac{e^{tgx}-1}{tgx-x}=\frac{e^{tg0}-1}{0-0}=\frac{0}{0}=[/m]

применяем правило Лопиталя

[m]lim_{x → 0}\frac{(e^{tgx}-1)`}{(tgx-x)`}=lim_{x → 0}\frac{e^{tgx}\cdot (tgx)`}{(tgx)`-(x)`}=lim_{x → 0}\frac{e^{tgx}\cdot \frac{1}{cos^2x}}{\frac{1}{cos^2x}-1}=[/m]

[m]=lim_{x → 0}\frac{e^{tgx}\cdot \frac{1}{cos^2x}}{\frac{1-cos^2x}{cos^2x}}=lim_{x → 0}\frac{e^{tgx}}{sin^2x}=(\frac{1}{0})= ∞ [/m]