Limit of \( \displaystyle \frac{\sin{\left(12 x + 4 \right)}}{9 x + 3} \) as \( x \to - \frac{1}{3} \)

Problem 1.87 · medium

1. \[ \lim_{x \to - \frac{1}{3}^+}\left(\frac{\sin{\left(12 x + 4 \right)}}{9 x + 3}\right) \]
   limit algebraStart with the original limit expression. Distribute the 3 in the denominator.✓ Proved
2. \[ = \lim_{x \to - \frac{1}{3}^+}\left(\frac{\frac{d}{d x} \sin{\left(12 x + 4 \right)}}{\frac{d}{d x} \left(9 x + 3\right)}\right) \]
   lhopitalThe limit is an indeterminate form 0/0, so apply L'Hopital's rule.✓ Proved
3. \[ = \lim_{x \to - \frac{1}{3}^+}\left(\frac{4 \cos{\left(12 x + 4 \right)}}{3}\right) \]
   simplifyCompute the derivatives of the numerator and denominator.✓ Proved
4. \[ = \frac{4}{3} \]
   limit simplify simplifyEvaluate the limit by substituting x = -1/3. Simplify the fraction and the argument of the cosine. Since cos(0) = 1, the final value is 4/3.✓ Proved