Limit of \( \displaystyle \frac{2 x - \sin{\left(2 x \right)}}{8 x^{3}} \) as \( x \to 0 \)

Problem 1.51 · medium

1. \[ \lim_{x \to 0^+}\left(\frac{2 x - \sin{\left(2 x \right)}}{8 x^{3}}\right) \]
   limitStart with the original limit.✓ Proved
2. \[ = \lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x - \sin{\left(2 x \right)}\right)}{\frac{d}{d x} 8 x^{3}}\right) \]
   lhopitalApply L'Hopital's rule because the limit is an indeterminate form 0/0.✓ Proved
3. \[ = \lim_{x \to 0^+}\left(\frac{2 - 2 \cos{\left(2 x \right)}}{24 x^{2}}\right) \]
   simplify factorCompute the derivatives. Factor out the constant 2.✓ Proved
4. \[ = \lim_{x \to 0^+}\left(\frac{1 - \cos{\left(2 x \right)}}{12 x^{2}}\right) \]
   simplifySimplify the fraction.✓ Proved
5. \[ = \lim_{x \to 0^+}\left(\frac{2 x^{2} + O\left(x^{4}\right)}{12 x^{2}}\right) \]
   rewriteUse the Taylor series expansion for cos(2x).✓ Proved
6. \[ = \lim_{x \to 0^+} \frac{1}{6} \]
   simplify algebra cancelKeep the leading term of the expansion. Expand the squared term. Cancel the x**2 terms.✓ Proved
7. \[ = \frac{1}{6} \]
   simplifyFinal evaluation.✓ Proved