Limit of \( \displaystyle \frac{\sin{\left(6 x + 3 \right)}}{2 x + 1} \) as \( x \to - \frac{1}{2} \)

Problem 1.234 · medium

1. \[ \lim_{x \to - \frac{1}{2}^+}\left(\frac{\sin{\left(6 x + 3 \right)}}{2 x + 1}\right) \]
   limit factorIdentify the limit to evaluate. Factor the argument of the sine function.✓ Proved
2. \[ = \lim_{x \to - \frac{1}{2}^+}\left(\frac{\frac{d}{d x} \sin{\left(6 x + 3 \right)}}{\frac{d}{d x} \left(2 x + 1\right)}\right) \]
   lhopitalApply L'Hopital's rule because the limit is an indeterminate form 0/0.✓ Proved
3. \[ = \lim_{x \to - \frac{1}{2}^+}\left(3 \cos{\left(6 x + 3 \right)}\right) \]
   simplify simplifyCompute the derivatives of the numerator and denominator. Simplify the fraction.✓ Proved
4. \[ = 3 \]
   limit simplify simplifyEvaluate the limit by substituting x = -1/2. Simplify the argument of the cosine function. Evaluate the final value.✓ Proved