Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 567: 103

$${\text{}}a = 1,{\text{ }}b = \pm 2{\text{ }}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{a - \cos bx}}{{{x^2}}} = 2 \cr & {\text{To apply L'Hopital's rule we need the result}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{a - \cos bx}}{{{x^2}}} = \frac{0}{0},{\text{ }}\,{\text{then}} \cr & a - \cos bx = 0{\text{ for }}\mathop {\lim }\limits_{x \to 0} \cos bx = 1,\,{\text{then}} \cr & a - 1 = 0 \cr & {\text{Let }}a = 1 \cr & \cr & {\text{Using L'Hopital's rule}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{a - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {a - \cos bx} \right]}}{{\frac{d}{{dx}}\left[ {{x^2}} \right]}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - \left( { - b\sin bx} \right)}}{{2x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{b\sin bx}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{b^2}\cos bx}}{2} \cr & {\text{Where }}\mathop {\lim }\limits_{x \to 0} \frac{{a - \cos bx}}{{{x^2}}} = 2,{\text{ then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{b^2}\cos bx}}{2} = 2 \cr & \mathop {\lim }\limits_{x \to 0} {b^2}\cos bx = 4 \cr & {b^2}\mathop {\lim }\limits_{x \to 0} \cos bx = 4 \cr & {b^2}\left( 1 \right) = 4 \cr & b = \pm 2, \cr & \cr & {\text{Then, }}a = 1,{\text{ }}b = \pm 2{\text{ }} \cr} $$