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

Problem 1.176 · medium

1. \[ \lim_{x \to - \frac{1}{3}^+}\left(\frac{e^{6 x + 2} - 1}{\sin{\left(3 x + 1 \right)}}\right) \]
   limitIdentify the limit to evaluate.✓ Proved
2. \[ = \lim_{x \to - \frac{1}{3}^+}\left(e^{6 x + 2} - 1\right) \left(\lim_{x \to - \frac{1}{3}^+} \sin{\left(3 x + 1 \right)}\right)^{-1} \]
   limit-lawCheck the limit of the numerator and denominator separately.Not checked
3. \[ = \tilde{\infty} \lim_{x \to - \frac{1}{3}^+}\left(e^{6 x + 2} - 1\right) \]
   limitThe denominator approaches zero.Not checked
4. \[ = \lim_{x \to - \frac{1}{3}^+} \text{NaN} \]
   limitThe numerator also approaches zero, creating an indeterminate form.Not checked
5. \[ = \lim_{x \to - \frac{1}{3}^+}\left(\frac{\frac{d}{d x} \left(e^{6 x + 2} - 1\right)}{\frac{d}{d x} \sin{\left(3 x + 1 \right)}}\right) \]
   lhopitalApply L'Hopital's Rule because the limit is of the form 0/0.Not checked
6. \[ = \lim_{x \to - \frac{1}{3}^+}\left(\frac{2 e^{6 x + 2}}{\cos{\left(3 x + 1 \right)}}\right) \]
   simplify algebraCompute the derivatives of the numerator and denominator. Simplify the constant coefficient.✓ Proved
7. \[ = 2 \]
   substitution simplify simplify limitSubstitute the value x = -1/3 into the expression. Simplify the exponents and trigonometric arguments. Evaluate the exponential and trigonometric functions. Final result.✓ Proved