Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 317: 38

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$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin x}}{{\sin x - x}} \cr & {\text{Use the quotient property of limits}} \cr & = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {\sin x - x} \right)}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{{{\left( 0 \right)}^2}\sin \left( 0 \right)}}{{\sin \left( 0 \right) - \left( 0 \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 & \frac{{\mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {\sin x - x} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{x^2}\sin x} \right]}}{{\frac{d}{{dx}}\left[ {\sin x - x} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\cos x + 2x\sin x}}{{\cos x - 1}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{{{\left( 0 \right)}^2}\cos \left( 0 \right) + 2\left( 0 \right)\sin \left( 0 \right)}}{{\cos \left( 0 \right) - 1}} = \frac{0}{{1 - 1}} = \frac{0}{0} \cr & {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\cos x + 2x\sin x}}{{\cos x - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{x^2}\cos x + 2x\sin x} \right]}}{{\frac{d}{{dx}}\left[ {\cos x - 1} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}\sin x + 2x\cos x + 2x\cos x + 2\sin x}}{{ - \sin x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}\sin x + 4x\cos x + 2\sin x}}{{ - \sin x}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{ - {{\left( 0 \right)}^2}\sin \left( 0 \right) + 4\left( 0 \right)\cos \left( 0 \right) + 2\sin \left( 0 \right)}}{{ - \sin \left( 0 \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[ { - {x^2}\sin x + 4x\cos x + 2\sin x} \right]}}{{\frac{d}{{dx}}\left[ { - \sin x} \right]}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}\cos x - 2x\sin x - 4x\sin x + 4\cos x + 2\cos x}}{{ - \cos x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}\cos x - 6x\sin x + 6\cos x}}{{ - \cos x}} \cr & {\text{Evaluate the limit when }}x \to 0 \cr & = \frac{{ - {{\left( 0 \right)}^2}\cos \left( 0 \right) - 6\left( 0 \right)\sin \left( 0 \right) + 6\cos \left( 0 \right)}}{{ - \cos \left( 0 \right)}} \cr & = \frac{6}{{ - 1}} = - 6 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin x}}{{\sin x - x}} = - 6 \cr} $$