Формула Тейлора (примеры)
Пример 6
Вычислить предел
y = sin(sin x)
y′ = cos(sin x)cos x
y″ = −sin(sin x)cos2 x − cos(sin x)sin x
y′′′ = −cos(sin x)cos3 x + 3sin(sin x)cos x sin x − cos(sin x)cos x
y(IV) = sin(sin x)cos4 x + 6cos(sin x)cos2 x sin x −
−3sin(sin x)sin2 x + 4sin(sin x)cos2 x + cos(sin x)sin x
y(V) = cos(sin x)cos5 x − 10sin(sin x)cos3 x sin x −
−15cos(sin x)cos x sin2 x + 10cos(sin x)cos3 x −
−15sin(sin x)sin x cos x + cos(sin x)cos x
y(0) = 0, y′(0) = 1, y″(0) = 0, y′′′(0) = −2, y(IV)(0) = 0, y(V)(0) = 12