Первый замечательный предел:
\infty } \frac{sin \frac{1}{ \alpha } }{ \frac{1}{ \alpha } } = 1" alt=" lim_{x - > \infty } \frac{sin \frac{1}{ \alpha } }{ \frac{1}{ \alpha } } = 1" align="absmiddle" class="latex-formula">
Поэтому:
\infty }(x + 1)sin \frac{5}{x + 1} = lim_{x - > \infty } \frac{5}{5} (x + 1)sin \frac{5}{x + 1} = \\ \\ lim_{x - > \infty } 5\frac{sin \frac{5}{ x + 1 } }{ \frac{5}{ x + 1 } } = 5 \times 1 = 5" alt=" lim_{x - > \infty }(x + 1)sin \frac{5}{x + 1} = lim_{x - > \infty } \frac{5}{5} (x + 1)sin \frac{5}{x + 1} = \\ \\ lim_{x - > \infty } 5\frac{sin \frac{5}{ x + 1 } }{ \frac{5}{ x + 1 } } = 5 \times 1 = 5" align="absmiddle" class="latex-formula">