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

Problem 1.357 · medium

1. \[ \lim_{x \to -1^+}\left(\frac{x - \sin{\left(x + 1 \right)} + 1}{\left(x + 1\right)^{3}}\right) \]
   limitEvaluate the limit of the function at x = -1.✓ Proved
2. \[ = \lim_{x \to -1^+}\left(\frac{\frac{d}{d x} \left(x - \sin{\left(x + 1 \right)} + 1\right)}{\frac{d}{d x} \left(x + 1\right)^{3}}\right) \]
   lhopitalApply L'Hopital's Rule because the limit is an indeterminate form 0/0.✓ Proved
3. \[ = \lim_{x \to -1^+}\left(\frac{1 - \cos{\left(x + 1 \right)}}{3 \left(x + 1\right)^{2}}\right) \]
   simplifyCompute the derivatives in the numerator and denominator.✓ Proved
4. \[ = \lim_{u \to 0^+}\left(\frac{1 - \cos{\left(u \right)}}{3 u^{2}}\right) \]
   substitutionUse the substitution u = x + 1, which implies u approaches 0 as x approaches -1.✓ Proved
5. \[ = \lim_{u \to 0^+}\left(\frac{\frac{d}{d u} \left(1 - \cos{\left(u \right)}\right)}{\frac{d}{d u} 3 u^{2}}\right) \]
   lhopitalApply L'Hopital's Rule to the new limit.✓ Proved
6. \[ = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{6 u}\right) \]
   simplifyCompute the derivatives.✓ Proved
7. \[ = \lim_{u \to 0^+}\left(\frac{\frac{d}{d u} \sin{\left(u \right)}}{\frac{d}{d u} 6 u}\right) \]
   lhopitalApply L'Hopital's Rule again.✓ Proved
8. \[ = \lim_{u \to 0^+}\left(\frac{\cos{\left(u \right)}}{6}\right) \]
   simplifyCompute the derivatives.✓ Proved
9. \[ = \frac{1}{6} \]
   limitEvaluate the final limit.✓ Proved