limit as x approaches-3+of-(sin(c-3x))/(e^{5x+15)-1}

解

x \to - 3 + lim (- \frac{sin ( c - 3 x )}{e ^{5 x + 15} - 1})

\infty

解答ステップ

x \to - 3 + lim (- \frac{sin ( c - 3 x )}{e ^{5 x + 15} - 1})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - x \to - 3 + lim (\frac{sin ( c - 3 x )}{e ^{5 x + 15} - 1})

\frac{sin ( c - 3 x )}{e ^{5 x + 15} - 1} = sin (c - 3 x) \frac{1}{e ^{5 x + 15} - 1}

= - x \to - 3 + lim (sin (c - 3 x) \frac{1}{e ^{5 x + 15} - 1})

x \to a lim [f (x) \cdot g (x)] = x \to a lim f (x) \cdot x \to a lim g (x)

不定形式の場合を除く

= - x \to - 3 + lim (sin (c - 3 x)) \cdot x \to - 3 + lim (\frac{1}{e ^{5 x + 15} - 1})

x \to - 3 + lim (sin (c - 3 x)) = sin (c + 9)

x \to - 3 + lim (\frac{1}{e ^{5 x + 15} - 1}) = \infty

= - sin (c + 9) \cdot \infty

簡素化

- sin (c + 9) \cdot \infty : \infty

= \infty