Limit of \( \displaystyle \frac{\cos{\left(2 x - 1 \right)} - 1}{e^{2 x - 1} - 1} \) as \( x \to \frac{1}{2} \)

Problem 1.456 · medium

1. \[ \lim_{x \to \frac{1}{2}^+}\left(\frac{\cos{\left(2 x - 1 \right)} - 1}{e^{2 x - 1} - 1}\right) \]
   limitStart with the original limit.✓ Proved
2. \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{\frac{d}{d x} \left(\cos{\left(2 x - 1 \right)} - 1\right)}{\frac{d}{d x} \left(e^{2 x - 1} - 1\right)}\right) \]
   lhopitalApply L'Hôpital's rule because the limit is of the form 0/0.✓ Proved
3. \[ = \lim_{x \to \frac{1}{2}^+}\left(- e^{1 - 2 x} \sin{\left(2 x - 1 \right)}\right) \]
   simplify simplify rewriteCompute the derivatives. Cancel the common factor of 2. Rewrite the fraction using a negative exponent.✓ Proved
4. \[ = \lim_{u \to 0^+}\left(- e^{- u} \sin{\left(u \right)}\right) \]
   substitutionSubstitute u = 2*x - 1, so u approaches 0 as x approaches 1/2.✓ Proved
5. \[ = \lim_{u \to 0^+}\left(- u e^{- u}\right) \]
   rewriteUse the small angle approximation sin(u) ~ u near 0.✓ Proved
6. \[ = \lim_{u \to 0^+}\left(- u\right) \lim_{u \to 0^+} e^{- u} \]
   limit-lawApply the product rule for limits.✓ Proved
7. \[ = 0 \]
   limit limitEvaluate the individual limits. Final result.✓ Proved