SOLUTION 8: $$\displaystyle{ \lim_{x \rightarrow 0} \frac{ \sin x^2 }{ x \tan x } } = \displaystyle{ \frac{ \sin(0)^2 }{ (0)\tan(0) } } = \displaystyle{ \frac{ (0)^2 }{ (0)(0) } } =\frac{"0"}{0}$$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately. Use the Chain Rule on the top. Use the Product Rule on the bottom.)
$$= \displaystyle{ \lim_{x \rightarrow 0} \frac{ 2x \cos x^2 }{ x \sec^2 x + (1) \tan x} } = \displaystyle{ \frac{ 2 (0) \cos (0)^2 }{ (0) \sec^2 (0) + \tan(0)} } = \displaystyle{ \frac{ 2 (0) (1)^2 }{ (0)(1)^2 + (0)} } = \displaystyle{ \frac{ 0 }{ (0)(1)^2 + (0)} } = \displaystyle{ \frac{"0"}{0} }$$ (Apply Theorem 1 for l'Hopital's Rule again. Differentiate the top using the Product Rule and the Chain Rule. Differentiate the bottom using the Product Rule and the Chain Rule.) $$= \displaystyle{ \lim_{x \rightarrow 0} \frac{ 2x \cdot 2x (- \sin x^2) + (2) \cos x^2 } { x \cdot 2 \sec x \cdot \sec x \tan x + (1) \sec^2 x + \sec^2 x} }$$ $$= \displaystyle{ \lim_{x \rightarrow 0} \frac{ 2 \cos x^2 - 4x^2 \sin x^2 } { 2x \sec^2 x \tan x + 2 \sec^2 x } }$$ $$= \displaystyle{ \frac{ 2 \cos (0)^2 - 4(0)^2 \sin (0)^2 } { 2(0) \sec^2 (0) \tan (0) + 2 \sec^2 (0) } }$$ $$= \displaystyle{ \frac{ 2 (1)^2 - 4(0)(0) } { 2(0) (1)^2 (0) + 2 (1)^2 } }$$ $$= \displaystyle{ \frac{ 2 - 0 } { 0 + 2 } }$$ $$= \displaystyle{ \frac{ 2 } { 2 } }$$ $$= 1$$