Limit of \( \displaystyle \frac{\sin{\left(3 x + 3 \right)}}{x + 1} \) as \( x \to -1 \)

Problem 1.307 · medium

1. \[ \lim_{x \to -1^+}\left(\frac{\sin{\left(3 x + 3 \right)}}{x + 1}\right) \]
   limit algebraStart with the original limit expression. Factor out 3 from the argument of the sine function.✓ Proved
2. \[ = \lim_{x \to -1^+}\left(\frac{\frac{d}{d x} \sin{\left(3 x + 3 \right)}}{\frac{d}{d x} \left(x + 1\right)}\right) \]
   lhopitalThe limit is an indeterminate form of type 0/0, so apply L'Hôpital's rule.✓ Proved
3. \[ = \lim_{x \to -1^+}\left(3 \cos{\left(3 x + 3 \right)}\right) \]
   simplifyCompute the derivatives of the numerator and denominator.✓ Proved
4. \[ = 3 \]
   substitution simplifySubstitute the value x = -1 into the expression. Evaluate the cosine of zero.✓ Proved