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

Problem 1.500 · medium

1. \[ \lim_{x \to 1^+}\left(\frac{x - \sin{\left(x - 1 \right)} - 1}{\left(x - 1\right)^{3}}\right) \]
   limit algebraIdentify the limit to evaluate. Rewrite the numerator by grouping terms.✓ 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'Hôpital'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.✓ Proved
4. \[ = \lim_{u \to 0^+}\left(1 - \frac{\cos{\left(u \right)}}{3 u^{2}}\right) \]
   substitutionSubstitute u = x - 1, which implies u approaches 0 as x approaches 1.Not checked
5. \[ = \lim_{u \to 0^+}\left(\frac{- \frac{u^{4}}{24} + \frac{u^{2}}{2} + 1}{3 u^{2}}\right) \]
   rewriteUse the Taylor series expansion for cos(u) near 0.Not checked
6. \[ = \lim_{u \to 0^+}\left(\frac{- \frac{u^{4}}{24} + \frac{u^{2}}{2}}{3 u^{2}}\right) \]
   simplifySimplify the numerator.Not checked
7. \[ = \lim_{u \to 0^+}\left(\frac{1}{6} - \frac{u^{2}}{72}\right) \]
   factor cancelFactor out u**2 from the numerator. Cancel the common u**2 term.✓ Proved
8. \[ = \lim_{u \to 0^+} \frac{1}{6} \]
   simplifyEvaluate the remaining constant limit.✓ Proved
9. \[ = \frac{1}{6} \]
   limitFinal result.✓ Proved