Задача №1136
Условие
Найти предел \(\lim_{h\to{0}}\frac{\sin(a+2h)-2\sin(a+h)+\sin{a}}{h^2}\).
Решение
\[
\lim_{h\to{0}}\frac{\sin(a+2h)-2\sin(a+h)+\sin{a}}{h^2}
=\left[\frac{0}{0}\right]
=\lim_{h\to{0}}\frac{2\sin(a+h)\cos{h}-2\sin(a+h)}{h^2}=\\
=\lim_{h\to{0}}\frac{2\sin(a+h)\cdot\left(\cos{h}-1\right)}{h^2}
=\lim_{h\to{0}}\left(-\frac{\sin^2\frac{h}{2}}{\left(\frac{h}{2}\right)^2}\cdot\sin(a+h)\right)
=-\sin{a}.
\]
Ответ:
\(-\sin{a}\)