Задача №1137
Условие
Найти предел \(\lim_{h\to{0}}\frac{\tg(a+2h)-2\tg(a+h)+\tg{a}}{h^2}\).
Решение
\[
\lim_{h\to{0}}\frac{\tg(a+2h)-2\tg(a+h)+\tg{a}}{h^2}
=\left[\frac{0}{0}\right]
=\lim_{h\to{0}}\frac{\tg(a+2h)-\tg(a+h)+\tg{a}-\tg(a+h)}{h^2}=\\
=\lim_{h\to{0}}\left(\frac{\sin{h}}{h^2\cos(a+h)}\cdot\left(\frac{1}{\cos(a+2h)}-\frac{1}{\cos{a}}\right)\right)
=\lim_{h\to{0}}\left(\frac{\sin{h}}{h^2\cos(a+h)}\cdot\frac{\cos{a}-\cos(a+2h)}{\cos{a}\cdot\cos(a+2h)}\right)=\\
=\lim_{h\to{0}}\left(\left(\frac{\sin{h}}{h}\right)^2\cdot\frac{2\sin(a+h)}{\cos{a}\cdot\cos(a+h)\cdot\cos(a+2h)}\right)
=\frac{2\sin{a}}{\cos^3{a}}.
\]
Ответ:
\(\frac{2\sin{a}}{\cos^3{a}}\)