SOLUTION 7: $$\displaystyle{ \lim_{x \rightarrow 0} \frac{ \arcsin 4x }{ \arctan 5x } } = \displaystyle{ \frac{ \arcsin(0) }{ \arctan(0) } } = \frac{"0"}{0}$$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately. Use the Chain Rule on the top and the bottom.)
$$= \displaystyle{ \lim_{x \rightarrow 0} \frac{ \frac{ 1}{ \sqrt{1-(4x)^2} } (4) }{ \frac{ 1 }{ 1+ (5x)^2 } (5) } }$$ $$= \displaystyle{ \lim_{x \rightarrow 0} \frac{ 4}{ \sqrt{1-16x^2} } \cdot \frac{ 1+ 25x^2 }{5} }$$ $$= \displaystyle{ \frac{ 4 }{ 5 } \cdot \frac{ 1+25(0)^2}{ 1-16(0)^2 } }$$ $$= \displaystyle{ \frac{ 4 }{ 5 } \cdot \frac{ 1+0}{ 1-0 } }$$ $$= \displaystyle{ \frac{4}{5} }$$