Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 317: 42

$\frac{1}{6}$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x\sin \left( {{x^2}} \right)}} \cr & {\text{Use the quotient property of limits}} \cr & = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {x - \sin x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {x\sin \left( {{x^2}} \right)} \right)}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{0 - \sin 0}}{{0\sin \left( {{0^2}} \right)}} = \frac{0}{0} \cr & {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr & {\text{the L'Hospital's Rule}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x\sin \left( {{x^2}} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {x - \sin x} \right]}}{{\underbrace {\frac{d}{{dx}}\left[ {x\sin \left( {{x^2}} \right)} \right]}_{{\text{product rule}}}}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{2{x^2}\cos \left( {{x^2}} \right) + \sin \left( {{x^2}} \right)}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{1 - \cos \left( 0 \right)}}{{2{{\left( 0 \right)}^2}\cos \left( {{0^2}} \right) + \sin \left( {{0^2}} \right)}} = \frac{0}{0} \cr & {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{2{x^2}\cos \left( {{x^2}} \right) + \sin \left( {{x^2}} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {1 - \cos x} \right]}}{{\underbrace {\frac{d}{{dx}}\left[ {2{x^2}\cos \left( {{x^2}} \right)} \right]}_{{\text{product rule}}} + \frac{d}{{dx}}\left[ {\sin \left( {{x^2}} \right)} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{0 + \sin x}}{{ - 4{x^3}\sin \left( {{x^2}} \right) + 4x\cos \left( {{x^2}} \right) + 2x\cos \left( {{x^2}} \right)}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{\sin 0}}{{ - 4{{\left( 0 \right)}^3}\sin \left( {{0^2}} \right) + 4\left( 0 \right)\cos \left( {{0^2}} \right) + 2\left( 0 \right)\cos \left( {{0^2}} \right)}} = \frac{0}{0} \cr & {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {\sin x} \right]}}{{\frac{d}{{dx}}\left[ { - 4{x^3}\sin \left( {{x^2}} \right) + 4x\cos \left( {{x^2}} \right) + 2x\cos \left( {{x^2}} \right)} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {\sin x} \right]}}{{\frac{d}{{dx}}\left[ { - 4{x^3}\sin \left( {{x^2}} \right)} \right] + \frac{d}{{dx}}\left[ {6x\cos \left( {{x^2}} \right)} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{ - 8{x^4}\cos \left( {{x^2}} \right) - 12{x^2}\sin \left( {{x^2}} \right) - 12x\sin \left( {{x^2}} \right) + 6\cos \left( {{x^2}} \right)}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{\cos \left( 0 \right)}}{{ - 8{{\left( 0 \right)}^4}\cos \left( 0 \right) - 12{{\left( 0 \right)}^2}\sin \left( 0 \right) - 12\sin \left( 0 \right) + 6\cos \left( 0 \right)}} \cr & = \frac{1}{6} \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{x\sin \left( {{x^2}} \right)}} = \frac{1}{6} \cr} $$